Contents · 2016-12-16 · ST334 ACTUARIAL METHODS version 2016/03 These notes are for ST334...

ST334 ACTUARIAL METHODSversion 2016/03

These notes are for ST334 Actuarial Methods. The course covers Actuarial CT1 and some related financialtopics.

Actuarial CT1 is called ‘Financial Mathematics’ by the Institute of Actuaries. However, we reserve the term‘financial mathematics’ for the study of stochastic models of derivatives and financial markets which is con-siderably more advanced mathematically than the material covered in this course.

There are 2 versions of these notes—one is in monochrome and is more suitable for printing and the othercontains bookmarks and hyperlinks and is intended for viewing in a PDF reader such as SumatraPDF.

Contents1 Simple and Compound Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1 Simple interest 1. 1.2 Compound interest 3. 1.3 Exercises 8. 1.4 Discount rate and discountfactor 10. 1.5 Exercises 15. 1.6 Time dependent interest rates 16. 1.7 Exercises 18.

2 Cash Flows, Equations of Value and Project Appraisal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.1 Cash flows 21. 2.2 Net present value and discounted cash flow 22. 2.3 Exercises 27. 2.4 Equa-tions of value and the internal rate of return 30. 2.5 Comparing two projects 33. 2.6 Exercises 35.2.7 Measuring investment performance 38. 2.8 Exercises 39.

3 Perpetuities, Annuities and Loans . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

3.1 Perpetuities 43. 3.2 Annuities 44. 3.3 Exercises 51. 3.4 Loan schedules 57. 3.5 Loans repaidby specified instalments 61. 3.6 Exercises 62.

4 Basic Financial Instruments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

4.1 Markets, interest rates and financial instruments 69. 4.2 Fixed interest government borrow-ings 70. 4.3 Other fixed interest borrowings 71. 4.4 Investments with uncertain returns 73.4.5 Derivatives 74. 4.6 Exercises 79.

5 Bonds, Equities and Inflation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

5.1 Bond calculations 81. 5.2 Exercises 87. 5.3 Equity calculations 90. 5.4 Real and money ratesof interest 91. 5.5 Exercises 93.

6 Interest Rate Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

6.1 Spot rates, forward rates and the yield curve 101. 6.2 Exercises 107. 6.3 Vulnerability tointerest rate movements 111. 6.4 Exercises 116. 6.5 Stochastic interest rate problems 123. 6.6 Ex-ercises 125.

7 Arbitrage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

7.1 Arbitrage 131. 7.2 Forward contracts 133. 7.3 Exercises 138.

Appendix 1: Some Mathematical Prerequisites . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

Appendix 2: Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

Exercises 1.3 145. Exercises 1.5 147. Exercises 1.7 148. Exercises 2.3 149. Exercises 2.6 152.Exercises 2.8 155. Exercises 3.3 158. Exercises 3.6 166. Exercises 4.6 173. Exercises 5.2 174.Exercises 5.5 179. Exercises 6.2 187. Exercises 6.4 193. Exercises 6.6 202. Exercises 7.3 206.

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PrologueOrder of Topics.

There are arguments for considering the formulæ for perpetuities and annuities covered in chapter 3 before thediscussion of project appraisal in chapter 2. This has not been done in order to avoid starting with a long listof formulæ on nominal and effective interest rates and annuities. However, this does mean that the solutionsof some of the problems on project appraisal in exercises 6 of chapter 2 (page 35) can be simplified by usingresults from section 2 of chapter 3 (page 44).

Exercises

An enormous number of exercises is included. You are not expected to do them all—but, it is very importantthat you practice solving exercises quickly and accurately. Only you can decide how many you need to do inorder to achieve the required proficiency to pass the tests and the examination.

BooksReferences.

Brealey, R & S. Myers (2005, eighth edition) Principles of Corporate Finance McGraw HillBrett, M. (2003, fifth edition) How to Read the Financial Pages Random HouseDonald, D.W.A.(1975) Compound Interest and Annuities-Certain HeinemannMcCutcheon, J.J. & W.F. Scott (1986) An Introduction to the Mathematics of Finance ButterworthRoss, S.M. (2002, second edition) An Elementary Introduction to Mathematical Finance CUP (Chapter 4)Steiner, R. (2007, second edition) Mastering Financial Calculations Prentice HallVaitilingam, R. (2001, fourth edition) The Financial Times Guide to Using the Financial Pages Prentice Hall

Most of the exercises have been taken from past examination papers for the Faculty and Institute of Actuar-ies. The web reference is http://www.actuaries.org.uk/students/pages/past-exam-papers and they arecopyright The Institute and Faculty of Actuaries.

Please send corrections and suggestions for improvement to [email protected].

Index of Notationsymbol page meaning

a∞ or a∞ ,i 43 §1.1 present value of perpetuitya∞ or a∞ ,i 43 §1.1 value at time 1 of a perpetuitya(m)∞ and a(m)

∞ 52 §3 (22) value at time 0 and at time t = 1/m of a perpetuity payable m timesper year

an or an,i 44 §2.1 present value of an annuity with n paymentsan or an,i 44 §2.1 value at time 1 of an annuity with n paymentsa(m)n or a(m)

n,i 45 §2.3 value at time 0 of an annuity with m payments per year for n yearsa(m)n or a(m)

n,i 45 §2.3 value at time 1/m of an annuity with m payments per year for n yearsa∞ 49 §2.7 value at time 0 of a continuously payable perpetuity with ρ(t) = 1an 49 §2.7 value at time 0 of a continuously payable annuity with ρ(t) = 1 for

t ∈ (0, n)k|an and k|an 47 §2.4 annuity payable annually for n years and delayed by k years

k|a(m)n and k|a

(m)n 47 §2.4 annuity payable m times per year for n years and delayed by k years

k|an 49 §2.7 value at time 0 of a continuously payable annuity with ρ(t) = 1 fort ∈ (k, k + n)

ACT/365 1 §1.4 actual number of days divided by 360A(p) 4 §2.6 accumulated value at time p of an investment of 1 at time 0

C 81 §1.1 redemption value of a bondCD 69 §1.3 certificate of depositc(i) 113 §3.5 convexityCP 69 §1.3 commercial paper

d = 1− ν 10 §4.1 discount rated(m) 12 §4.5 nominal discount rate payable m times per yeardp 12 §4.5 nominal discount rate payable per time period, p(Da)n 49 §2.6 present value of a decreasing annuity for n yearsDCF 22 §2.1 discounted cash flowd(i) 113 §3.3 effective duration or volatility or modified durationdM (i) 111 §3.2 duration or Macaulay duration or discounted mean termDPP 33 §5.3 discounted payback periodδ = ln(1 + i) 6 §2.8 nominal rate of interest compounded continuously or

force of interest corresponding to the effective annual interest, iδ : (0,∞)→ [0,∞) 17 §6.1 force of interest function

f 81 §1.1 face or par value of a bondFRA 76 §5.5 forward rate agreementfT,k 103 §1.5 forward interest rate p.a. for money at time T in the future for length

of time kFT,k = ln(1 + fT,k) 103 §1.6 forward force of interestft,T,k 104 §1.7 forward interest rate p.a. at time t for investment starting at time T

for length of time kFt,T,k = ln(1 + ft,T,k) 105 §1.7 forward force of interestFt,T = limk→0 Ft,T,k 105 §1.7 instantaneous forward rate at time t for investment starting at time T

g = fr/C 83 §1.5 annual coupon divided by redemption value of a bond

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i 1 §1.2 simple interesti 3 §2.1 compound interestiep 4 §2.5 the effective interest rate over the period pi(m) = i1/m 4 §2.5 nominal rate of interest compounded m times per yeariM and iR 91 §4.1 money and real rate of interestip 4 §2.5 nominal rate of interest per period p(Ia)n and (Ia)n 48 §2.5 value at time 0 and at time 1 of an increasing annuity for n years(Ia)n 50 §2.8 increasing continuously payable annuity with a step function ρ(Ia)n 50 §2.8 increasing continuously payable annuity with ρ(t) = t for t ∈ (0, n)(Is)n and (Is)n 48 §2.5 value at time n and at time n + 1 of an increasing annuity for n years(Is)n 50 §2.8 increasing continuously payable annuity with a step function ρ(Is)n 50 §2.8 increasing continuously payable annuity with ρ(t) = t for t ∈ (0, n)IRR 30 §4.1 internal rate of return

LIBID 70 §1.4 London Inter-Bank Bid RateLIBOR 70 §1.4 London Inter-Bank Offered RateLIMEAN 70 §1.4 average of LIBOR and LIBIDLIRR 38 §7.3 linked internal rate of return

MWRR 38 §7.1 money weighted rate of return

NPV 22 §2.1 net present valueν = 1/(1 + i) 10 §4.1 discount factorν : [0,∞)→ [0,∞) 17 §6.2 present value function

P (n, i) 81 §1.1 current price of abond with n years to maturity and yield i

r 81 §1.1 coupon rate per year of a bondρ : (0,∞)→ R 25 §2.5 continuous payment stream

sn or sn,i 45 §2.2 value at time n of an annuity with n paymentssn or sn,i 45 §2.2 value at time n + 1 of an annuity with n paymentss(m)n or s(m)

n,i 45 §2.3 value at time n of an annuity with m payments per year for n yearss(m)n or s(m)

n,i 45 §2.3 value at time n + 1/m of an annuity with m payments per year forn years

sn 49 §2.7 value at time n of a continuously payable annuity with ρ(t) = t fort ∈ (0, n)

TWRR 38 §7.1 time weighted rate of return

yt 101 §1.2 t-year spot rate of interestYt = ln(1 + yt) 101 §1.2 t-year spot force of interest

CHAPTER 1

Simple and Compound Interest

1 Simple interest

1.1 Background. The existence of a market where money can be borrowed and lent allows people to transferconsumption between today and tomorrow.

The amount of interest paid depends on several factors, including the risk of default and the amount of depre-ciation or appreciation in the value of the currency.

Interest on short-term financial instruments is usually simple rather than compound.

1.2 The simple interest problem over one time unit. This means there is one deposit (sometimes called theprincipal) and one repayment. The repayment is the sum of the principal and interest and is sometimes calledthe future value.

Suppose an investor deposits c0 at time t0 and receives a repayment c1 at the later time t0 + 1. Then the gain(or interest) on the investment c0 is c1 − c0. The interest rate, i is given by

i =c1 − c0

c0

Note that

c1 = c0(1 + i) and c0 =1

1 + ic1

The quantity c0 is often called the present value (at time t0) of the amount c1 at time t0 + 1. The last formulaencapsulates the maxim “a pound today is worth more than a pound tomorrow”.

Example 1.2a. Suppose the amount of £100 is invested for one year at 8% per annum. Then the repayment inpounds is

c1 = (1 + 0.08)× 100 = 108The present value in pounds of the amount £108 in one year’s time is

c0 =1

1 + 0.08× 108 = 100

Of course, the transaction can also be viewed from the perspective of the person who receives the money. Aborrower borrows c0 at time t0 and repays c1 at time t0 + 1. The interest that the borrower must pay on the loanis c1 − c0.

1.3 The simple interest problem over several time units. Consider the same problem over n time units,where n > 0 is not necessarily integral and may be less than 1. The return to the investor will be

cn = c0 + nic0 = c0(1 + ni)

It follows that present value (at time t0) of the amount cn at time t0 + n is

c0 =1

1 + nicn

1.4 Day and year conventions. Interest calculations are usually based on the exact number of days as aproportion of a year.Domestic UK financial instruments usually assume there are 365 days in a year—even in a leap year. Thisconvention is denoted ACT/365 which is short for actual number of days divided by 365.

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Example 1.4a. Assuming ACT/365, how much interest will the following investments pay?(a) A deposit of £1000 for 60 days at 10% p.a.(b) A deposit of £1000 for one year which is not a leap year at 10% p.a.(c) A deposit of £1000 for one year which is a leap year at 10% p.a.

Solution. (a) 1000× 0.1× 60/365 = 16.44. (b) 1000× 0.1 = 100. (c) 1000× 0.1× 366/365 = 100.27.

Under ACT/365, a deposit at 10% for one year which is a leap year will actually pay slightly more than 10%—itwill pay:

10%× 366365

= 10.027%

Most overseas money markets assume a year has 360 days and this convention is denoted ACT/360.

1.5 The time value of money. In general, an amount of money has different values on different dates.Suppose an amount c0 is invested at a simple interest rate of i. The time value is represented in the followingdiagram:

............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... ............................

........

........................

past value present value future value..............................................................

..............................................................

..............................................................

c0

(1 + in)c0 c0(1 + in)

︷︸︸︷︷︸︸︷n years n years

A calculation concerning simple interest can only be done with respect to one fixed time point. This fixed timepoint is sometimes called the focal point or valuation date.

Here is an example of an incorrect value obtained by using more than one focal point:

Consider the amount c0 at today’s time t0 and simple interest i p.a. This amount c0 had value x = c0/(1 + 2i)two years ago at time t0 − 2. But the value x at time t0 − 2 has value x(1 + i) = c0(1 + i)/(1 + 2i) one yearlater at time t0 − 1. This argument is incorrect because two different focal points (t0 − 2 and t0 − 1) havebeen used. In fact, the amount c0 at time t0 had value c0/(1 + i) at time t0 − 1.

Simple interest means that interest is not given on interest—hence using more than one focal point will give anerror.

Example 1.5a. A person takes out a loan of £5000 at 9% p.a. simple interest. He repays £1000 after 3 months (timet0 + 1/4), £750 after a further two months (time t0 + 5/12) and a further £500 after a further month (time t0 + 1/2).What is the amount outstanding one year after the loan was taken out (time t0 + 1)?

Solution. The question asks for the amount outstanding at time t0 + 1. Hence we must take t0 + 1 as the focal point.This gives

5000[1 + 0.09]− 1000[

1 + 0.09(

912

)]− 750

[1 + 0.09

(7

12

)]− 500

[1 + 0.09

(12

)]= 3070.625

Here is another solution: £1,000 is borrowed for 3 months, £750 for 5 months, £500 for 6 months and £2,750 for12 months. Hence the total interest payment is[

1000× 0.0912× 3]

+[

750× 0.0912× 5]

+[

500× 0.0912× 6]

+ [2750× 0.09] = 320.625

The total capital outstanding after 12 months is £2,750 and hence the total amount due is £2,750 + £320.625 =£3,070.625. The issuer of the loan would round this up to £3,070.63.

1.6

Summary.• Terminology: simple interest; time value of money; present value; ACT/365• Simple interest formulæ:

cn = (1 + ni)c0 and c0 =1

1 + nicn

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2 Compound interest2.1 Explanation of compound interest. Compound interest means that interest is given on any interestwhich has already been credited.

Example 2.1a. Suppose an investor invests £720 for two years at a compound interest rate of 10% p.a. Find theamount returned.

Solution. The amount in pounds is c2 = 720(1 + i)2 = 720× 1.12 = 871.20.

In general, if c0 denotes the initial investment, i denotes the interest rate p.a. and cn denotes the accumulatedvalue after n years, then

cn = c0(1 + i)n

The accumulated interest is clearly c0[(1 + i)n − 1]. This formula assumes the interest rate is constant andhence does not depend on the time point or the amount invested.

2.2 Simple versus compound interest. With simple interest, the growth of the accumulated value is linear;with compound interest, the growth is exponential. Exercise 25 shows that for the investor, simple is preferableto compound for less than 1 year whilst compound is preferable to simple for more that 1 year.

Accounts which allow investors to withdraw and reinvest will pay compound interest—otherwise investorswould continually withdraw and reinvest.

2.3 Present value. Suppose t1 ≤ t2 then an investment of c/(1 + i)t2−t1 at time t1 will produce a return of cat time t2. It follows that

c

(1 + i)t2−t1is the discounted value at time t1 of the amount c at time t2.In particular, the expression “the present value (time 0) of the amount c at time t” means exactly the same as“the discounted value at time 0 of the amount c at time t.”Thus, if the compound interest rate is i per unit time, then the present value of the amount c at time t is

c

(1 + i)t= cνt where ν =

11 + i

is called the discount factor.

Example 2.3a. Inflation adjusted rate of return. Suppose the inflation rate is i0 per annum. Suppose further that aninvestment produces a return of i1 per annum. What is the inflation adjusted rate of return, ia, of the investment?In particular, if i1 = 0.04 (or 4% per annum) and the inflation rate is i0 = 0.02 (or 2% per annum), find ia.

Solution. Suppose the capital at time 0 is c0. After 1 year, the accumulated value will be c0(1 + i1). Adjusting forinflation, the time 0 value of the amount c0(1 + i1) at time 1 is

c0(1 + i1)(1 + i0)

Hence

ia =1 + i11 + i0

− 1 =i1 − i01 + i0

Clearly, if i0 is small then ia ≈ i1 − i0.For our particular case ia = 0.02/1.02 = 0.0196 or 1.96%. The usual approximation is i1 − i0 which gives 2%.

2.4 Nominal and effective rates of interest—a numerical example. Loan companies often quote a ratesuch as 15% per annum compounded monthly. This means that a loan of £100 for one year will accumulate to

£100(

1 +0.1512

)12

= £100× 1.1608 = £116.08.

The effective interest rate per annum is 16.08%.

Similarly, suppose an investment pays a rate of interest of 12% p.a. compounded 12 times per year. Then thecorresponding effective annual rate of interest is defined to be the rate which would produce the same amountof interest per year; i.e. (

1 +0.1212

)12

− 1 = (1 + 0.01)12 − 1 = 0.1268 or 12.68%.


Example 2.4a. Suppose £1000 is invested for two years at 10% per annum compounded half-yearly. What is theamount returned?

Solution. The amount is £1000× (1.05)4 = £1215.51.

2.5 Nominal and effective rates of interest—the general case. In general, i(m) and i1/m denote the nominalrate of interest per unit time which is compoundedm times per unit time and which corresponds to the effectiveinterest rate of i per unit time. The relation between the two quantities is given by:

i =(

1 +i(m)

m

)m− 1 (2.5a)

The quantity i(m) = m[(1 + i)1/m − 1

]is also described as a nominal rate of interest per unit time paid mthly,

or convertible mthly, or with mthly rests. It is left as an exercise to prove that

i = i(1) > i(2) > i(3) > · · ·If p = 1/m, the quantity i(m) is also denoted ip and is called the nominal rate of interest per period p corre-sponding to the effective interest rate of i per unit time. Writing equation (2.5a) in terms of p gives

(1 + i)p = 1 + pip where p =1m

and ip = i(m) (2.5b)

The quantity iep = pip = i(m)/m is called the effective interest rate over the period p. In particular

i(1) = i = ie1

is simply called the effective interest rate per unit time.

Clearly, equivalent effective annual interest rates should be used for comparing nominal interest rates withdifferent compounding intervals.

Example 2.5a. Suppose the nominal rate of interest is 10% p.a. paid quarterly. Find(a) the effective annual rate of interest; (b) the effective rate of interest over 3 months.

Solution. In this case p = 1/4 and ip = 0.1. The answer to part (b) is iep = 0.1/4 = 0.025. Hence the effectiveannual rate is i = (1 + iep)4 − 1 = (1 + 0.025)4 − 1 = 1.1038− 1 which is 10.38%.

2.6 Terminology. Here are two special cases:• if the rate of interest offered on a 6 month investment is 10%, then this means that the investor will receive5% of his capital in interest at the end of the 6 months.In general, for a situation of less than one year, the equivalent simple interest rate per annum is usually quoted.• a 3 year investment at 10% means that the investor will receive an amount of interest of c0×1.13 where c0 ishis initial investment. This assumes the interest can be reinvested. In this case, the quoted rate is the one yearrate which is compounded.

The basic time unit is always assumed to be one year unless it is explicitly specified otherwise. Thus thephrase “a 3 year investment at 10%” used above is just shorthand for “a 3 year investment at 10% p.a.”

In general, to specify an interest rate, we need to specify the basic time unit and the conversion period. Aninterest rate is called effective if the basic time unit and the conversion period are identical—and then interestis credited at the end of the basic time period. When the conversion period does not equal the basic time period,the interest rate is called nominal. Two more examples:• An effective interest rate of 12% over 2 years is the same as a nominal interest rate of 6% p.a. convertibleevery 2 years. There is one interest payment after 2 years and this payment has size equal to 12% of thecapital1.• An effective interest rate of 3% per quarter is the same as a nominal interest rate of 12% p.a. payablequarterly.

In general, for both p ∈ (0, 1) and p > 1, an effective interest rate iep over the period p is also described as anominal interest rate of ip = iep/p per annum convertible 1/p times a year.

1 In practice, this is not quite the same as 6% p.a. simple interest because the simple interest received after year one couldbe invested elsewhere for year two.


Calculations should usually be performed on effective interest rates—the nominal rate is just a matter of pre-sentation. Remember that:

nominal rate = frequency× effective rate or i(m) = m× iep.For many problems, you should follow the following steps (or possibly a subset of the following steps):• From ip, the nominal rate for period p, calculate iep, the effective rate for period p, by using nominalrate = frequency × effective rate (where frequency is 1/p).• Convert to the effective rate for period q by using (1 + ieq)1/q = (1 + iep)1/p.• Convert to the nominal rate for period q by using nominal rate = frequency × effective rate wherefrequency is 1/q.

For p ≥ 0, the accumulation factor2 A(p) = (1 + i)p = 1 + pip gives the accumulated value at time p of aninvestment of 1 at time 0.

2.7 Worked examples on nominal and effective interest rates.

Example 2.7a. What is the simple interest rate per annum which is equivalent to a nominal rate of 9% compoundedsix-monthly, if the money is invested for 3 years?

Solution. Now £1 invested at a nominal rate of 9% compounded six-monthly will accumulate to (1.045)6 = 1.302 in3 years. If the annual rate of simple interest is r then we need

1 + 3r = (1.045)6

and so r = 0.100753 giving a simple interest rate of 10.08%.

Example 2.7b. Suppose the effective annual interest rate is 41/2%. What is the nominal annual interest rate payable(a) quarterly; (b) monthly; (c) every 2 years?

Solution. (a) Now (1 + iep)4 = 1 + i = 1.045 implies iep = 0.011065 for one quarter. Hence answer (a) is i(4) = 4iep =0.04426 or nominal 4.426% p.a. payable quarterly. (b) (1 + iep)12 = 1.045 implies i(12) = 12iep = 0.04410. Henceanswer (b) is nominal 4.410% p.a. payable monthly. (c) Finally, (1 + iep)1/2 = 1.045 implies iep = 0.0920. Henceanswer (c) is ip = iep/2 = 0.0460, or nominal 4.6% p.a. payable every 2 years.

Example 2.7c. An investor invests and withdraws the following amounts in a savings account at the stated timesdate investment withdrawal

1 April 2000 £1,5001 April 2001 £1,0001 April 2002 £1,8001 April 2003 £1,500

Between 1 April 2000 and 1 April 2001, the account paid nominal 6% per annum convertible quarterly. Between1 April 2001 and 1 October 2002, the account paid 5% per annum effective. From 1 October 2002, the account paid4% per annum effective. Interest is added to the account after close of business on 31 March each year. Find theaccumulated amount at 1 October 2003.

Solution. There are two possible approaches.• Calculate the amount in the account at every time point where there is a transaction.At 31 March 2001, amount in the account is 1,500 × 1.0154. Hence at the end of the day 1 April 2001, the amountis 1,500× 1.0154 − 1,000. Proceeding in this manner gives the answer

(((1500× 1.0154 − 1000)× 1.05 + 1800)× 1.051/2 × 1.041/2 + 1500)× 1.041/2 = 4110.41 (2.7a)• Convert every amount into the value at 1 October 2003 and then add these amounts together.So the deposit on 1 April 2000 is worth 1,500 × 1.0154 × 1.053/2 × 1.04 on 1 October 2003. The withdrawal on1 April 2001 is worth − 1,000 × 1.053/2 × 1.04 on 1 October 2003. And so on. This leads to the same answer. Infact, the second approach is just the same as expanding out equation (2.7a) in the first approach.

2 The term factor is used because the accumulated value is c0A(p) if the initial capital is c0. Some books use A(t1, t2) todenote the accumulated value at time t2 of an investment made at time t1. We then haveA(0, t2) = A(0, t1)A(t1, t2) and so

A(t1, t2) =A(0, t2)A(0, t1)

=A(t2)A(t1)

Hence there are no real advantages in using this alternative notation A(t1, t2) and we stick with A(t).


Example 2.7d. Suppose a financial instrument pays a nominal interest rate of 21/2% p.a. convertible every 2 years.(a) What is the effective interest rate over 2 years. (b) What is the equivalent effective annual interest rate?

Solution. (a) The effective interest rate over 2 years is 5%. (b) Let i denote the equivalent effective annual interestrate. Then (1 + i)2 = 1.05 and so i = 0.0247. Hence the effective annual interest rate is 2.47%.

Example 2.7e. The nominal interest rates for local authority deposits on a particular day are as follows:

Term Nominal interest rate (%)

overnight 103/16

one month 93/43 months 97/86 months 10

Find the accumulation of £1000 for (a) one month and (b) 3 months.

Solution. For (a), we have1000(1 + 0.0975/12) = 1008.12

For (b), we have1000(1 + 0.09875/4) = 1024.69

Note that the one month nominal rate corresponds to an effective annual rate of(1 +

0.097512

)12

− 1 = 0.1019772 or 10.20%.

The 3 month nominal rate corresponds to an effective annual rate of(1 +

0.098754

)4

− 1 = 0.1024674 or 10.25%.

One reason for this apparent inconsistency is that the market allows for the fact that interest rates change over time.If both rates corresponded to the same effective annual rate of interest, then the 3 month rate could be obtained fromthe 1 month rate by doing the calculation:

4×

[(1 +

0.097512

)3

− 1

]= 0.09829433

which is slightly less than the rate of 97/8% in the table above.

Example 2.7f. Show that a constant rate of simple interest implies a declining effective rate of interest.

Solution. Let i denote the rate of simple interest per unit time. Then the effective rate of interest for time period n is:(1 + in)− (1 + i(n− 1))

1 + i(n− 1)=

i

1 + i(n− 1)which decreases with n. It is i for time period 1 and i/(1 + i) for time period 2, etc.

Example 2.7g. Show that a constant rate of compound interest implies a constant rate of effective interest.

Solution. Let i denote the rate of compound interest per unit time. Then the effective rate of interest for time periodn is:

(1 + i)n − (1 + i)n−1

(1 + i)n−1 =(1 + i)− 1

1= i

2.8 Continuous compounding: the force of interest. From equation (2.5a) we have

i(m) = m[(1 + i)1/m − 1

](2.8a)

Now limn→∞ n(x1/n − 1) = lnx if x > 1 (see exercise 22). Using this result on equation (2.8a) shows thatlimm→∞ i

(m) exists and equals ln(1 + i). Denote this limit by δ. Henceδ = ln(1 + i) or i = eδ − 1

where δ = limm→∞ i(m) = limp→0 ip is called the nominal rate of interest compounded continuously or the

force of interest corresponding to the effective annual rate of interest, i.

It follows that if an amount c0 is invested at a rate δ per annum compounded continuously, then after 1year it will have grown to c1 = c0e

δ and after p years it will have grown to c0epδ, where p is not necessarily

an integer.


The accumulation factor can be expressed in terms of δ, the force of interest, as follows:

A(p) = epδ and so δ =A′(p)A(p)

for all p ≥ 0

Example 2.8a. Suppose the interest rate on a 91 day £10,000 investment is: (a) simple 6% per annum;(b) nominal 6% per annum compounded daily; (c) nominal 6% per annum compounded continuously.Find the amount returned after 91 days. For all three cases, assume the year has 365 days.

Solution. For (a), the amount is £10,000(

1 + 0.06× 91365

)= £10,149.59.

For (b), it is £10,000(

1 +0.06365

)91

= £10,150.70.

For (c), we know that c0 grows to c0e0.06 after 1 year and hence grows to c0e

0.06×91/365 after 91 days. This has value£10,150.71.The answer to part (c) could be derived from first principles as follows: suppose each day is split into n pieces andthe interest is compounded after each piece. After 91 days, an initial principal of c0 grows to

ct = c0

(1 +

0.06365n

)91n

Using limn→∞

(1 +

x

n

)n= ex gives lim

n→∞ct = c0

(e0.06/365

)91= c0e

0.06×91/365 as above.

Example 2.8b. For the previous example, find the equivalent effective daily interest rates for (a), (b) and (c).

Solution. (a)(

1 + 0.06× 91365

)1/91

− 1 = 0.000163 (b)0.06365

= 0.000164 (c) e0.06/365 − 1 = 0.000164

Example 2.8c. Continuation of the previous example. Find the equivalent effective annual interest rates.

Solution. (a)(

1 + 0.06× 91365

)365/91

− 1 = 0.0614 (b)(

1 +0.06365

)365

− 1 = 0.0618 (c) e0.06 − 1 = 0.0618

Example 2.8d. Continuation of the previous example. Find the equivalent simple interest rates per annum.

Solution. (a) 0.06 (b)36591

[(1 +

0.06365

)91

− 1

]= 0.060446 (c)

36591

(e0.06×91/365 − 1) = 0.060451

The general formulæ are left as exercises.

Example 2.8e. Suppose a financial instrument with a maturity date of 13 weeks (or 91 days) pays an interest rateof 6.4% per annum. What is the equivalent nominal continuously compounded rate per annum? Assume the yearhas 365 days.

Solution. Recall that for situations of less than one year, the equivalent simple interest rate per annum is quoted.Hence, assuming an investment of £c0, the instrument pays the amount

£c0

(1 +

0.064× 91365

)after 91 days. The effective annual rate is

i1 =(

1 + 0.064× 91365

)365/91

− 1 = 0.0656

which is 6.56%. The nominal continuously compounded rate is

δ = ln(1 + i1) =36591× ln

(1 + 0.064× 91

365

)= 0.0635 which is 6.35%.

Alternatively, equating the accumulated value after 91 days gives

e91δ/365 = 1 + 0.064× 91365

Exercises 3 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed

2.9

Summary.• Compound interest formulæ:

cn = (1 + i)nc0 and c0 = νncn where ν =1

1 + iis the discount factor

• Nominal and effective interest rates. Nownominal rate = frequency× effective rate.

Hence if ip = i(1/p) denotes the nominal rate of interest for period p and iep denotes the effective interestrate for period p, then

i(m) = ip = miep where m = 1/p is the frequency.Hence i(m) denotes the nominal rate of interest compounded m times per unit time. If i denotes theeffective interest rate per unit time, then

1 + i =(

1 +i(m)

m

)mor (1 + i)p = 1 + iep

• Continuous compounding. Suppose the force of interest (or the nominal rate of interest compoundedcontinuously) is δ per annum. Then continuous compounding implies the amount c0 accumulates toc0e

kδ after k years.Also, if i denotes the equivalent effective annual rate of interest, then:

1 + i = eδ and δ = ln(1 + i)

3 Exercises (exs1-1.tex)

1. (a) Suppose a loan of £5000 is to be repaid by the amount £6000 at the end of two months. What is the annual rateof simple interest?

(b) How long will it take £5000 to earn £100 at 6% p.a. simple interest?

2. The simple yield on a deposit is 9.5% on the ACT/360 basis. What is the equivalent simple yield on the ACT/365basis?

3. Suppose £1000 is invested for two years at 12% p.a. compounded monthly. How much interest is earned (a) duringthe first year; (b) during the second year?

4. Suppose the interest rate quoted for a one year deposit is 10% with all the interest paid on maturity. What is theequivalent nominal annual interest rate convertible monthly?

5. Suppose the terms of a £1,000 loan are 12% p.a. convertible monthly. One way to repay the loan is £10 at the end ofeach month for 11 months with a final payment of £1010 at the end of the twelfth month. Instead of this repaymentregime, suppose no intermediate payments are made; what is the payment at the end of the twelfth month?

6. Suppose the simple interest rate quoted for a 13 week (91 day) investment under ACT/365 is 10% p.a. What is theeffective annual rate?

7. Six years ago, A borrowed £1,000 at an effective interest rate of 6% for 2 years. How much did A owe one year ago?

8. A bank pays £975 for a note which pays £1,000 in 6 month’s time. What is the effective annual interest rate?

9. Suppose a financial instrument pays an effective interest rate of 2% per 6 months. Find (a) the equivalent nominalinterest rate per annum payable every 6 months; (b) the equivalent effective annual interest rate; (c) the equivalenteffective interest rate payable quarterly.

10. (a) Find the effective annual interest rate equivalent to a nominal rate of 6% payable every 6 months. (b) Find thenominal interest rate p.a. payable every 6 months which is equivalent to an effective interest rate of 6% p.a.

11. Suppose the nominal interest rate p.a. payable quarterly is 8%. Find the equivalent nominal interest rate p.a. payableevery 6 months?

12. Consider the effective interest rate of 1% per month.(a) Find the equivalent effective interest rate (i) per 3 months; (ii) per 6 months; (iii) per annum; (iv) per 2 years.(b) Find the equivalent nominal interest rate p.a. (i) payable monthly (ii) payable every 3 months; (iii) payable

every 6 months; (iv) payable every year; (v) payable every 2 years.

1 Simple and Compound Interest Sep 27, 2016(9:51) Exercises 3 Page 9

13. Suppose an investment of £1,500 returns a total (principal plus interest) of £1,540 after 91 days under ACT/365.What is the simple interest rate on this investment? What is the effective annual rate of interest?

14. Suppose someone has 3 outstanding loans: loan 1 is cleared by a repayment of £300 in 2 years, loan 2 by a repaymentof £150 in 4 years and loan 3 by a repayment of £500 in 5 years. Suppose interest is being charged at a nominal rateof 5% p.a. payable every 6 months.(a) What single payment in 6 months’ time will pay off all 3 loans?(b) At what time could a repayment of £950 clear all 3 debts (to the nearest month)?

15. Suppose £50,000 is invested for 3 years at 5% per annum paid annually. When an interest payment is received, it isimmediately invested to the end of the 3 year period. Suppose that interest rates have risen to 6% p.a. at the end ofthe first year, and are at 5.5% p.a. at the end of the second year. What is the total amount received at the end of the 3years?

16. (a) A loan has a fixed nominal rate of 6% p.a. payable every 3 months. The loan is paid off by 20 payments of £800every 6 months for 10 years. Now suppose the loan is repaid by 10 equal payments at the end of each year; whatis the value of these annual payments?

(b) Now suppose the loan has a fixed nominal rate of i% p.a. payable every 3 months and the loan is paid off by 20payments of £x every 6 months for 10 years. If the loan is repaid by 10 equal payments of £y at the end of eachyear, express y in terms of x and i.

17. (a) What is the present value of £2,000 in 3 years’ time using 7% per annum?(b) What is the present value of £2,000 in 182 days’ time using 7% per annum? (Use simple interest and ACT/365.)

18. Suppose a bond pays a simple interest rate of 4.5% p.a. under ACT/360. What is the equivalent simple interest rateunder ACT/365?

19. (a) Suppose a deposit of £500 leads to a repayment of £600 after 4 years. If the same amount of £500 is depositedfor 6 years, what is the repayment?

(b) Suppose a loan of £5,500 leads to a repayment of £5,750 after one year.(i) If the loan is extended for one further year, calculate the new repayment.

(ii) If the term of the loan is reduced to 9 months, what should the repayment be?

20. (a) Suppose an investment for k days has a continuously compounded interest rate of r. Find i, the equivalentsimple annual interest rate.

(b) Suppose an investment for k days is quoted as having a simple interest rate of i. What is the equivalent contin-uous compounded rate?

21. Suppose the force of interest is δ = 0.12.(a) Find the equivalent nominal rate of interest per annum compounded

(i) every 7 days (assume 365 days in the year); (ii) monthly.(b) Suppose £1000 is invested for (i) 7 days (ii) one month. What is the amount returned?

22. Suppose x > 1. By using the inequality 1 < t1/n < x1/n for t ∈ (1, x) and n > 0, show that lnx < n(x1/n − 1) <x1/n lnx. Hence prove that limn→∞ n(x1/n − 1) = lnx for x > 1.

23. Suppose i ≥ 0 and f : (0,∞)→ R is defined by f (m) = i(m) = m[(1 + i)1/m − 1

]. Show that f ↓ as m ↑.

24. Suppose i denotes the effective annual interest rate and ip denotes the nominal interest rate per period p. Let δ denotethe force of interest corresponding to i. Show that

δ = limp↓0

ip = limp↓0

A(p)− 1p

25. Suppose A1 denotes the accumulation factor function for simple interest at rate i% per annum and A2 denotes theaccumulation function for compound interest at rate i% per annum. Hence A1(t) = 1 + ti and A2(t) = (1 + i)t.Prove that A1(t) > A2(t) for 0 < t < 1 and A2(t) > A1(t) for t > 1.For the investor, simple is preferable to compound for less than 1 year; compound is preferable to simple for morethat 1 year.

26. On some financial instruments, compound interest is used but with linear interpolation between integral time units.(a) Suppose A borrows £1,000 at 10% per annum compound interest. After 1 year and 57 days how much does he

owe? (Use ACT/365.)(b) Using linear interpolation between integral time units, how much does he owe?(c) Show that using linear interpolation between integral time units in this way is always unfavourable to the bor-

rower.


27. A 90-day treasury bill was bought by an investor for a price of £91 per £100 nominal. After 30 days the investor soldthe bill to a second investor for a price of £93.90 per £100 nominal. The second investor held the bill to maturitywhen it was redeemed at par.Determine which investor obtained the higher annual effective rate of return.

(Institute/Faculty of Actuaries Examinations, September 2004) [3]

28. An insurance company calculates the single premium for a contract paying £10,000 in 10 years’ time as the presentvalue of the benefit payable, at the expected rate of interest it will earn on its funds. The annual effective rate ofinterest over the whole of the next 10 years will be 7%, 8% or 10% with probabilities 0.3, 0.5 and 0.2 respectively.(a) Calculate the single premium. (b) Calculate the expected profit at the end of the contract.


29. Calculate the nominal rate of interest per annum convertible quarterly which is equivalent to:(a) an effective rate of interest of 0.5% per month.(b) a nominal rate of interest of 6% per annum convertible every 2 years.

(Institute/Faculty of Actuaries Examinations, September 2002) [2+2=4]

30. A 90-day government bill is purchased for £96 at the time of issue and is sold after 45 days to another investor for£97.50. The second investor holds the bill until maturity and receives £100.Determine which investor receives the higher rate of return.


31. An investor purchases a share for 769p at the beginning of the year. Halfway through the year he receives a dividend,net of tax, of 4p and immediately sells the share for 800p. Capital gains tax of 30% is paid on the difference betweenthe sale and the purchase price.Calculate the net annual effective rate of return the investor obtains on the investment.


32. (a) Calculate the time in days for £3,600 to accumulate to £4,000 at:(i) a simple rate of interest of 6% per annum;

(ii) a compound rate of interest of 6% per annum convertible quarterly;(iii) a compound rate of interest of 6% per annum convertible monthly.

(b) Explain why the amount takes longer to accumulate in (i)(a).(Institute/Faculty of Actuaries Examinations, September 2006) [4+1=5]

33. Calculate the time in days for £3,000 to accumulate to £3,800 at:(a) a simple rate of interest of 4% per annum; (b) a compound rate of interest of 4% per annum effective.

(Institute/Faculty of Actuaries Examinations, April 2015) [4]

34. An investor pays £120 per annum into a savings account for 12 years. In the first four years, the payments are madeannually in advance. In the second four years, the payments are made quarterly in advance. In the final four years,the payments are made monthly in advance.The investor achieves a yield of 6% per annum convertible half-yearly on the investment.Calculate the accumulated amount in the savings account at the end of 12 years.


4 Discount rate and discount factor4.1 Discount factor and discount rate over one time unit. Suppose the quantity c0 accumulates to c1 inone time unit. We already know that the quantity c1− c0 is called the interest on the deposit c0. It is also calledthe discount on the repayment c1.

The interest rate is i =c1 − c0

c0and the discount rate is d =

c1 − c0

c1=

i

1 + i

Thus c1 = c0(1 + i), c0 = c1(1− d) and ic0 = dc1. The last formula can be remembered in the form:

interest rate × present value = discount rate × future value.

Note thatc0 =

11 + i

c1 = νc1 where ν =1

1 + iis the discount factor.

Hence c0 is the present value (at time t0) of the amount c1 at time t0 + 1, and

present value = discount factor × future value


It is easy to confuse the terms ‘discount rate’ (denoted d) and ‘discount factor’ (denoted ν), but the contextusually provides some assistance. The term usually used in finance is discount rate. Note that a discount rate(like an interest rate) is usually expressed as a percentage. The key relations between d, ν and i are given by:

ν = 1− d d =i

1 + iν =

11 + i

d = iν

The formula d = iν has the following interpretation: a discount of d received at time 0 is the same as an interestpayment of i received at time 1 discounted back to time 0 value.

Example 4.1a. Suppose the amount of £100 is invested for one year at 8% per annum. Then the repayment inpounds is c1 = (1 + 0.08)× 100 = 108. The present value of the amount £108 in one year’s time is

c0 =1

1 + 0.08× 108 = 100

The discount factor is 1/1.08 = 25/27 ≈ 0.9259. The discount rate is 8/108 = 2/27 ≈ 0.074 or 7.4%.

4.2 Simple discount over several time units. If the amount cn is due after n time units and a simple discountrate of d per time unit applies, then the amount that needs to be invested now is c0 = (1−nd)cn. It follows thatif i denotes the simple interest rate per time unit, then

1− nd =1

1 + niand so i =

d

1− ndand d =

i

1 + ni

Example 4.2a. A commercial bill with a maturity date of September 30 and a maturity value of £100,000 is pricedon August 1 at 5% p.a. simple discount. Find the amount of the discount.

Solution. The number of days is 61. Using c0 = cn(1− nd) shows that the amount of discount is cnnd = 100,000×(61/365)× 0.05 = 835.62.The interest of £835.62 is gained in 61 days on a principal of £100,000 − £835.62. This leads to a simple interestrate per annum of

36561

(835.62

100,000− 835.62

)=

100,000× 0.05100,000− 835.62

=0.05

1− (61/365)× 0.05= 0.0504

Alternatively, using (1 + ni)(1− nd) = 1 gives i = d/(1− nd) = 0.0504.Note that the simple interest rate is not equal to the simple discount rate d = 0.05.

4.3 Discount instrument. The coupon is the interest payment made by the issuer of a security. It is based onthe face value or nominal value. The face value is generally repaid (or redeemed) at maturity, but sometimes itis repaid in stages. The coupon amounts are calculated from the face value.

The yield is the interest rate which can be earned on an investment. This is implied by the current market pricefor an investment—as opposed to the coupon which is based on the coupon rate and face value.

A discount instrument is a financial instrument which does not carry a coupon. Because there is no interest,these are sold at less than the face value—in other words, they are sold at a discount. Treasury bills are sold inthis way. Because there is only one payment, simple interest is used for the calculations.

Example 4.3a. A Treasury bill for £10 million is issued for 90 days. Find the selling price in order to obtain a yieldof 10%.

Solution. Using cn = c0(1 + ni) shows that the bill must be sold for

c0 =cn

1 + ni=

10,000,0001 + (90/365)× 0.1

= 9,759,358.29

The term maturity proceeds means the same as face value. The term market price denotes the present value.


4.4 The key formulæ for simple interest and simple discount are cn = (1 + ni)c0 and c0 = (1− nd)cn. Otherresults are simple consequences of these two formulæ.Some books state the results in words as follows:

mathematical formula formula in words

c0 =cn

1 + nimarket price =

maturity proceeds1 + (yield× length of investment)

i =d

1− ndyield =

discount rate1− (discount rate× length of investment)

d =i

1 + nidiscount rate =

yield1 + (yield× length of investment)

c0 = cn(1− nd) market price = face value× (1− discount rate× length of investment)cn − c0 = cndn amount of discount = face value× discount rate× length of investment

Example 4.4a. A Treasury bill for £10 million is issued for 90 days. If the yield is 10%, what is the amount of thediscount and what is the discount rate?

Solution. The key formulæ are cn = (1 + ni)c0 and c0 = (1− nd)cn. In this question we are given cn, n and i; hencewe can find the required quantities c0 and d. Here is one way: from (1− nd)(1 + ni) = 1 we get

d =i

1 + ni=

0.11 + 0.1× 90/365

=365

3740= 0.09759 or 9.76% p.a.

Hence the amount of the discount is

cn − c0 = ndcn =90365× 365

3740× 10,000,000 =

90,000,000374

= 240,641.71

4.5 Effective and nominal discount rates. Most financial instruments use compound interest and the termseffective and nominal also apply to discount rates.Suppose d(m) denotes the nominal discount rate per annum payable m times per year. This is equivalent to theeffective discount rate d(m)/m per 1/m of a year. Hence it is equivalent to the effective discount d per annumwhere

1− d =(

1− d(m)

m

)mExample 4.5a. The effective discount rate is 1% per month. Find the equivalent nominal discount rate per annum.

Solution. It is 12% per annum convertible monthly.

If p = 1/m, the quantity d(m) is also denoted dp and is called the nominal discount rate payable per timeperiod p. This is equivalent to the effective discount rate dep = d(m)/m per 1/m of a year. As for interest rates,we have the relations d = d(1) = de1 and so 1− d = (1− dep)1/p.

An effective discount rate of dep over p years is described as a nominal annual discount rate of dep/p payableevery p years (if p ≥ 1), or as a nominal annual discount rate of dep/p convertible 1/p times per year ifp ∈ (0, 1).

The key results for nominal discount rates are:

nominal = effective× frequency

1− d =(

1− d(m)

m

)m= (1− dep)1/p

(1 +

i(m)

m

)(1− d(m)

m

)= (1 + iep)(1− dep) = 1

For many problems, the procedure is as follows (or possibly a subset of the following steps):

• From the nominal rate for period p, calculate dep, the effective rate for period p by using nominal rate= effective rate × frequency, where frequency is 1/p.• Convert to the effective rate for period q by using (1− deq)1/q = (1− dep)1/p.• Convert to the nominal rate for period q by using nominal rate = effective rate × frequency wherefrequency is 1/q.

Example 4.5b.(a) The nominal interest rate is 6% per annum payable quarterly. What is the effective interest rate per annum?(b) The nominal discount rate is 6% per annum payable quarterly. What is the effective discount rate per annum?


Solution. (a) Nominal 6% p.a. payable quarterly is equivalent to an effective rate of 11/2% for a quarter, which isequivalent to 1.0154− 1 = 0.06136, or 6.136% per annum. (b) Nominal discount rate 6% p.a. payable quarterlyis equivalent to an effective discount rate of 11/2% for a quarter. If d denotes the effective annual discount rate then1− d = (1− 0.015)4 = 0.9854. Hence we have an effective discount rate of 5.866% p.a.

Example 4.5c. Show that1d(m) =

1m

+1i(m) and d(m) = m

[1− (1 + i)−1/m

]Hence deduce that

limm→∞

d(m) = limm→∞

i(m) = δ where δ = ln(1 + i).

Solution. From the relation(1 +

i(m)

m

)(1− d(m)

m

)= 1 we get d(m) =

mi(m)

i(m) +mand hence the first relation. For the second relation use 1 + i(m)/m = (1 + i)1/m. Now i(m) → δ as m → ∞ by theresult in section 2.8. Using this on the first relation gives the final required result.

Example 4.5d. Suppose the effective interest rate is 1% per month.(a) Find the effective discount rate (i) per 3 months; (ii) per annum; (iii) per 2 years.(b) Find the nominal discount rate (i) payable every 3 months; (ii) payable annually; (iii)payable every 2 years.

Solution. (a) The effective interest rates are (i) 1.013−1 per 3 months, (ii) 1.0112−1 p.a. (iii) 1.0124−1 per 2 years.Hence the values of dep are: (i) 1− (1/1.01)3; (ii) 1− (1/1.01)12; (iii) 1− (1/1.01)24 by using (1 + iep)((1−dep) = 1.(b) The values of d(m) = m× dep are: (i) 4(1− (1/1.01)3); (ii) 1− (1/1.01)12; (iii) 0.5(1− (1/1.01)24).

4.6 Relation between interest payable mthly in advance and in arrear.

Example 4.6a. Suppose someone borrows 1 at time 0 for repayment at time 1 at an annual rate of interest of iwhere i > 0.• Suppose further that the borrower repays the interest by equal instalments of x at the end of each mth subinterval(i.e. at times 1/m, 2/m, 3/m, . . . , 1− 1/m, 1). Find x.

Solution. Let ν = 1/(1 + i) denote the value at time 0 of the amount 1 at time 1. The cash flow is as follows:

Time 0 1m

2m · · · m−1

m 1Cash Flow −1 x x · · · x 1 + x

We have

1 =m∑i=1

xνi/m +1

1 + i= x

ν1/m(1− ν)1− ν1/m + ν

Now i > 0 implies ν < 1 and so

x =1− ν1/m

ν1/m = (1 + i)1/m − 1 =i(m)

m= iep where p =

1m

Alternatively, the payments of x at times 1/m, 2/m, 3/m, . . . , 1 should be equivalent to a single payment of i attime 1. Hence we want x to satisfy x

(ν1/m + ν2/m + · · · + ν

)= i/(1 + i) = 1− ν = d.

Hence i(m) is the total interest paid for the loan schedule above and i(m)/m is the size of each of them payments.This provides an alternative definition of i(m).

• Now suppose that the borrower repays the interest by equal instalments of y at the start of each mth subinterval(i.e. at times 0, 1/m, 2/m, 3/m, . . . , (m− 1)/m). Find y.

Solution. The cash flow is now:Time 0 1

m2m · · · m−1

m 1Cash Flow −1 + y y y · · · y 1

In this case,

1 =m−1∑i=0

yνi/m +1

1 + i= y

1− ν1− ν1/m + ν

and so


y = 1− ν1/m = 1− (1− d)1/m =d(m)

m= dep where p = 1/m.

Alternatively, the payments of y at times 0, 1/m, 2/m, . . . , (m− 1)/m should be equivalent to a single payment ofi at time 1. Hence we want y to satisfy y

(1 + ν1/m + ν2/m + · · · + ν(m−1)/m

)= i/(1 + i) = 1− ν = d.

Hence d(m) is the total interest paid for the loan schedule above and d(m)/m is the size of each of the mpayments. This provides an alternative definition of d(m).

If the annual effective interest rate is i and d is given by (1− d)(1 + i) = 1, then the preceding example showsthat the 4 series of payments in the following table (4.6a) are equivalent.

Time 01m

2m

· · · m− 2m

m− 1m

1

d

d(m)

m

d(m)

m

d(m)

m· · · d(m)

m

d(m)

m

i(m)

m

i(m)

m· · · i(m)

m

i(m)

m

i(m)

mi

(4.6a)

4.7 Continuous compounding: the force of interest. Recall (see section 2.8) that 1 + i = eδ for continuouscompounding where δ denotes the force of interest per annum and i denotes the equivalent effective annualrate of interest.Hence, if the force of interest is δ per annum and the investment c0 receives continuously compounded interestfor n days, then the amount returned is c0e

δn/365. Hence the discount factor is e−δn/365.The general result is dep = 1− e−pδ; this follows from iep = epδ − 1 and (1 + iep)(1− dep) = 1.

Example 4.7a. Suppose an investment of 91 days has an interest rate of 6% per annum continuously compounded.What is the corresponding simple interest rate per annum? What is the 91 day discount factor?

Solution. The corresponding simple interest rate per annum is:

i =36591

(e0.06×91/365 − 1

)= 0.06045 which is 6.045%.

Now c1 = c0e0.06×91/365. Hence the discount factor is

e−0.06×91/365 = 0.98516

4.8

Summary.• Discount rate (d) and discount factor (ν):

ν =1

1 + id = 1− ν (1 + i)(1− d) = 1

• Simple discount over several time units.

cn = (1 + ni)c0 c0 = (1− nd)cn (1 + ni)(1− nd) = 1interest rate× present value = discount rate× future value (ic0 = dcn)

• Discount instrument: only the face value is paid on maturity—there is no coupon. They are sold forless than the face value—i.e. they are sold at a discount.• Coupon and yield: The coupon is based on the face value and the yield is based on the market value.• Compound interest.

ν = 1− d =(

1− d(m)

m

)m= (1− dep)1/p where dep =

d(m)

mand p =

1m

.(1 +

i(m)

m

)(1− d(m)

m

)= (1 + iep)(1− dep) = 1



1. If the simple interest rate on a 91 day investment is 8%, what is the discount factor?

2. Suppose the interest rate quoted on a 3 year investment is 8% per annum compounded annually. What is the 3 yeardiscount factor?

3. Suppose the interest rate is 81/3% p.a. How much would you pay for a note repaying £3,380 in 2 years’ time?

4. Suppose £720 is invested for 2 years at a compound interest rate of i = 1/6 per annum.(a) What is the amount returned after 2 years? (b) What is the discount rate per annum?

5. Suppose the size of an initial investment at time 0 is c0 = 275 and the discount rate is 331/3% per annum. Find thevalue of the repayment c1 after one year.

6. Suppose the interest rate is increased in the simple interest problem. Show that the discount rate is also increasedand the discount factor is reduced.

7. Consider the simple interest problem. Suppose the interest rate is increased by k times where k > 1. Prove that thediscount rate is not increased by as much as k times.

8. Assume i > 0. Prove the following relations:

(a) i > d (b) iν = d (c) id = i− d (d)i2

1 + i=

d2

1− d= id

9. (a) Suppose the nominal discount rate is 21/2% p.a. payable every 2 years. What is the effective annual discountrate per annum?

(b) Suppose the effective interest rate is 41/2% per annum. (i) What is the equivalent nominal annual discount ratepayable quarterly? (ii) What is the equivalent nominal annual discount rate payable monthly? (iii) What is theequivalent nominal annual discount rate payable every 3 years?

10. Suppose the effective rate of discount dep with period p is equivalent to the effective rate of interest ieq with periodq. Show that

(1 + ieq)1/q(1− dep)1/p = 1

11. Suppose the interest rate is 10% per annum.(a) Find the effective interest rate for a period of (i) k months; (ii) p years.(b) Find the effective discount rate for a period of (i) k months; (ii) p years.

12. Suppose the nominal discount rate is 8% payable every 3 months. What is the equivalent nominal discount ratepayable every 6 months?

13. Suppose an amount c0 is invested at time 0.(a) How many years does it take for the accumulated value to double at a rate i per annum compounded annually?(b) How many years does it take for the accumulated value to double at a nominal rate i per annum compounded

continuously.(c) Calculate the numerical answers when i is 1%, 5%, 7% and 10%.(d) Suppose the answer to part (a) is denoted by na and the answer to part (b) is denoted by nb. Show that

nb = ln 2/i < na < ln 2/(i(1− i/2)) for i > 0.

14. (a) Suppose the amount c0 is invested for 2 years at the simple accumulated interest rate of i2 for the 2 years (i.e.c0 grows to c0(1 + i2) after 2 years; the rate i2 is not per annum). Suppose the corresponding compound interestrate is i1 per annum. (i) Show that i2 > 2i1. (ii) Show that d2 < 2d1 where d2 is the simple accumulateddiscount rate and d1 is the annual discount rate.

(b) Suppose t = 2s and the amount c0 is invested for t years at the simple accumulated interest rate of it for thet years. Let i1 denote the corresponding annual compound interest rate. Now suppose the same amount c0 isinvested for s years at that simple accumulated interest rate is which corresponds to the same annual compoundinterest rate i1. Show that it > 2is and dt < 2ds.

15. (a) Suppose the discount rate is 1% for one month. Find the discount rate for (i) 2 months; (ii) k months.(b) Suppose a bank has an effective discount rate of 2% for a duration of 3 months. How much will the bank pay

for a 9 month note repaying £5,000?

16. Let δ = ln(1 + i).(a) Show that i(m) = m(eδ/m − 1) and d(m) = m(1− e−δ/m).(b) Show that i > i(2) > i(3) > · · · and d < d(2) < d(3) < · · ·.(c) Show that limm→∞ i(m) = limm→∞ d(m) = δ.


17. A government issues a 90-day Treasury Bill at a simple rate of discount of 6% per annum. Calculate the effectiverate of return per annum received by an investor who purchases the Bill at issue and holds it to maturity.


18. An investment is discounted for 28 days at a simple rate of discount of 4.5% per annum. Calculate the annualeffective rate of interest. (Institute/Faculty of Actuaries Examinations, April 2006) [3]

19. A government issues a 90-day Treasury Bill at a simple rate of discount of 5% per annum. Calculate the rate ofreturn per annum convertible half yearly received by an investor who purchases the bill and holds it to maturity. (UseACT/365.) (Institute/Faculty of Actuaries Examinations, September 2000) [3]

20. The rate of discount per annum convertible quarterly is 8%. Calculate:(a) The equivalent rate of interest per annum convertible half-yearly.(b) The equivalent rate of discount per annum convertible monthly.

(Institute/Faculty of Actuaries Examinations, April 2001) [2+2=4]

21. The rate of interest is 4.5% per annum effective.(i) Calculate

(a) the annual effective rate of discount;(b) the nominal rate of discount per annum convertible monthly;(c) the nominal rate of interest per annum convertible quarterly;(d) the effective rate of interest over a five year period.

(ii) Explain why your answer in part(i)(b) is higher than your answer to part (i)(a).(Institute/Faculty of Actuaries Examinations, September 2013) [5+2=7]

22. An investor is considering two investments. One is a 91-day deposit which pays a rate of interest of 4% per annumeffective. The second is a treasury bill.Calculate the annual simple rate of discount from the treasury bill if both investments are to provide the same effectiverate of return. (Institute/Faculty of Actuaries Examinations, September 2012) [3]

23. A 182-treasury bill, redeemable at $100, was purchased for $96.50 at the time of issue and later sold to anotherinvestor for $98 who held the bill to maturity. The rate of return received by the initial purchaser was 4% per annumeffective.

(i) Calculate the length of time in days for which the initial purchaser held the bill.(ii) Calculate the annual simple rate of return achieved by the second investor.

(iii) Calculate the annual effective rate of return achieved by the second investor.(Institute/Faculty of Actuaries Examinations, April 2015) [2+2+2=6]

24. An investor wishes to obtain a rate of interest of 3% per annum effective from a 91-day treasury bill. Calculate:(a) the price the investor must pay per £100 nominal;(b) the annual simple rate of discount from the treasury bill.


25. The nominal rate of discount per annum convertible monthly is 5.5%.(i) Calculate, giving all your answers as a percentage to three decimal places:

(a) the equivalent force of interest;(b) the equivalent effective rate of interest per annum;(c) the equivalent nominal rate of interest per annum convertible monthly.

(ii) Explain why the nominal rate of interest per annum convertible monthly calculated in part (i)(c) is less than theequivalent annual effective rate of interest calculated in part (i)(b).

(iii) Calculate, as a percentage to three decimal places, the effective annual rate of discount offered by an investmentthat pays £159 in eight years’ time in return for £100 invested now.

(iv) Calculate, as a percentage to three decimal places, the effective annual rate of interest from an investment thatpays 12% interest at the end of each two-year period.

(Institute/Faculty of Actuaries Examinations, September 2015) [3+1+1+1=6]

6 Time dependent interest rates6.1 Continuous compounding with time-dependent interest rate. Suppose an amount c0 is invested ata rate δ per annum compounded continuously. After k years it will have grown to c0e

kδ, where k is notnecessarily an integer.

Now suppose that the interest rate varies with time, t. In particular, suppose we have a value ip(t) for all p > 0and t ≥ 0 which gives the nominal interest rate per period p for an investment starting at time t. For t ≥ 0, let


A(t) denote the accumulated value at time t of an investment of 1 at time 0. HenceA(t + p) = A(t)[1 + pip(t)] for all p > 0.

Assuming A(t) is differentiable implies

limp↓0

ip(t) = limp↓0

A(t + p)−A(t)pA(t)

exists for all t > 0.

Denote this limit by δ(t). Hence

δ(t) =A′(t)A(t)

for all t > 0. (6.1a)

This leads to a function δ : (0,∞)→ [0,∞) which is called the force of interest function.Equation (6.1a) gives

d

dtlnA(t) = δ(t)

and hence

A(t2) = A(t1) exp(∫ t2

t1

δ(s) ds)

for t2 ≥ t1 (6.1b)

Using A(0) = 1 gives

A(t) = exp(∫ t

0δ(s) ds

)for t ≥ 0. (6.1c)

Example 6.1a. Suppose that t = t1 + · · · + tn and the force of interest is constant. Show thatA(t) = A(t1) · · ·A(tn)

Solution. Just use A(t) = exp(∫ t

0 δ ds)

= eδt. This result is true for ordinary constant compound interest.

6.2 Present and discounted value. If we have continuous compounding with time dependent force of inter-est δ(t), then the present value of the amount c at time t > 0 is

c exp(−∫ t

0δ(s) ds

)Similarly if t2 > t1, then the discounted value at time t1 of the amount c at time t2 is

c exp(−∫ t2

t1

δ(s) ds)

The formal development proceeds as follows: we postulate a function δ : (0,∞) → [0,∞) which will becalled the force of interest function. Then

Definition 6.2a. The present value function corresponding to the force of interest function, δ, is thefunction ν : [0,∞)→ [0,∞) defined by

ν(t) = exp(−∫ t

0δ(s) ds

)(6.2a)

and so ν(t) is the present value (time 0) of the amount 1 at time t.

Using A(t) = 1/ν(t) gives equation (6.1c) and hence equation (6.1b).Note that ν(t2)/ν(t1) = A(t1)/A(t2) is the discounted value at time t1 of the amount 1 at time t2.

Example 6.2b. Express ip(t), the nominal rate of interest per period p, in terms of(a) δ(t), the force of interest function. (b) ν(t), the present value function.

Solution. (a) Using A(t + p) = A(t)[1 + pip(t)] and equation (6.1c) gives

ip(t) =exp

(∫ t+pt

δ(s) ds)− 1

p(b) Using A(t + p) = A(t)[1 + pip(t)] and A(t) = 1/ν(t) gives

ip(t) =1p

[ν(t)

ν(t + p)− 1]


6.3 Further results. For further results on the present value of a mixture of discrete and continuous paymentstreams, see section 2 of the next chapter.

6.4

Summary.• Force of interest is a function δ : (0,∞)→ [0,∞).• The accumulated value at time t of the amount 1 at time 0 is

A(t) = exp(∫ t

0δ(s) ds

)for t > 0.

• The present value (time 0) of the amount 1 at time t is


0δ(s) ds

)


1. (a) Suppose £1,000 is invested for 6 months at a force of interest of 0.05 per annum. Find the accumulated value.(b) Now suppose £1,000 is invested for 6 months at an effective rate of interest of 5% p.a. Find the accumulated

value.

2. Suppose the force of interest function, δ, is given byδ(t) = 0.04 + 0.01t for t ≥ 0.

If an investment of £1,000 is made at time 1, find its value at time 6.

3. Suppose we have continuous compounding with time dependent interest rate δ : (0,∞) → R. Denote the presentvalue of the amount 1 at time t ≥ 0 by ν(t). Hence:


0δ(s) ds

)for t > 0.

(a) Given ν, how can δ be determined? (b) If ν(t) = e−bt, find δ(t).

4. Let δ(t) denote the average of δ(t) over the interval (0, t). Hence

δ(t) =1t

∫ t

0δ(s) ds for t > 0.

We also set δ(0) = δ(0). Clearly

A(t) = exp(t δ(t)

)Suppose the force of interest is the constant δ. Find A(t), ν(t) and δ(t).

5. Suppose the function δ : (0,∞) → R is nondecreasing. Prove that the function δ : (0,∞) → R, is also non-decreasing.

6. Suppose we have continuous compounding with time dependent interest rate δ : (0,∞) → R. Denote the presentvalue of the amount 1 at time t > 0 by ν(t). Hence:


0δ(s) ds

)for t > 0.

Prove that the function δ : (0,∞)→ R, is non-decreasing if and only ifν(αt) ≥

(ν(t)

)αfor all 0 ≤ α ≤ 1 and for all t > 0.

7. Suppose δ(t) = 0.06× 0.7t where time t is measured in years. Find the present value of £100 in 31/2 years’ time.

8. (a) Suppose the force of interest function is δ : (0,∞)→ R. Show that ip(t), the nominal interest rate per period pfor an investment starting at time t is given by:

ip(t) =1p

[exp

(∫ t+p

t

δ(u) du)− 1]

(b) Suppose that δ(t) = 0.06× 0.7t. Find ip(t) for p > 0 and t > 0.


9. Calculate the time in days for £1,500 to accumulate to £1,550 at(a) a simple rate of interest of 5% per annum(b) a force of interest of 5% per annum (Institute/Faculty of Actuaries Examinations, September 2005) [4]

10. The force of interest δ(t) at time t is a + bt2 where a and b are constants. An amount of £200 invested at time t = 0accumulates to £210 at time t = 5 and £230 at time t = 10.Determine a and b. (Institute/Faculty of Actuaries Examinations, September 2005) [5]

11. (a) Calculate the present value of £100 in ten years’ time at the following rates of interest/discount.(i) a rate of interest of 5% per annum convertible monthly

(ii) a rate of discount of 5% per annum convertible monthly(iii) a force of interest of 5% per annum

(b) A 91-day Treasury bill is bought for $98.91 and is redeemed at $100. Calculate the equivalent annual effectiverate of interest obtained from the bill. (Institute/Faculty of Actuaries Examinations, September 2005) [4+3=7]

12. The force of interest, δ(t), is a function of time and at any time t, measured in years, is given by the formula:

δ(t) =

0.04 + 0.01t if 0 ≤ t ≤ 8;0.07 if 8 ≤ t.

(a) Derive, and simplify as far as possible, expressions for A(t) where A(t) is the accumulated amount at time t ofan investment of £1 invested at time t = 0.

(b) Calculate the present value at time t = 0 of £100 due at time t = 10.(Institute/Faculty of Actuaries Examinations, April 2001) [5+2=7]

13. The force of interest δ(t) is a function of time and at any time t, measured in years, is given by the formula:

δ(t) =

0.05 0 ≤ t ≤ 30.09− 0.01t 3 < t ≤ 80.01t− 0.03 8 < t

(a) If £500 is invested at t = 2 and a further £800 is invested at t = 9, calculate the accumulated amount at t = 10.(b) Determine the constant effective rate of interest per annum, to the nearest 1% which would lead to the same

result as in (a) being obtained. (Institute/Faculty of Actuaries Examinations, April 2003) [7+3=10]

14. The nominal rate of discount per annum convertible quarterly is 8%.(i) Calculate the equivalent force of interest.

(ii) Calculate the equivalent effective rate of interest per annum.(iii) Calculate the equivalent nominal rate of discount per annum convertible monthly.

(Institute/Faculty of Actuaries Examinations, September 2012) [1+1+2=4]

15. The force of interest, δ(t), is a function of time and at any time t (measured in years) is given by

δ(t) = 0.05 + 0.02t for 0 ≤ t ≤ 5;

0.15 for t > 5(i) Calculate the present value of £1,000 due at the end of 12 years.

(ii) Calculate the annual effective rate of discount implied by the transaction in (i).(Institute/Faculty of Actuaries Examinations, September 2008) [5+2=7]

CHAPTER 2

Cash Flows, Equations of Valueand Project Appraisal

1 Cash Flows

1.1 Introduction. When assessing the value of an investment, it often helps to list the times and values ofany cash flows. The timings and values of these cash flows may be certain or uncertain.

Here are some examples.

Zero-coupon bond. This is a security which provides a specified cash amount at some specified future time.Suppose the investor purchases the bond for the price c0 and receives the lump sum S at time n. Then thecash-flow sequence can be represented by the following table:

Time 0 nCash flow −c0 S

Fixed-interest security. Suppose the maturity date is n and the investor receives a coupon (or interest payment)at times 1, 2, . . . , n and a lump sum S (called the redemption value) at time n. Then the cash flow is:

Time 0 1 · · · n− 1 nCash flow −c0 c · · · c c + S

Fixed-interest securities may be issued by governments, companies, local authorities and other public bodies.

Index-linked security. These are similar to fixed-interest securities, but the interest payments and possibly thefinal payments are linked to some index (for example, the RPI). This means that the sizes of the future cashflows are unknown at time 0, although they may be known in real terms.

Annuity. An investor pays a premium of c0 at time 0 and receives payments of c at times 1, 2, . . . , n.

Time 0 1 · · · n− 1 nCash flow −c0 c · · · c c

Equity. An ordinary shareholder receives dividends each year (or six-monthly). The size of the dividend isdetermined each year by the company.

Time 0 1 2 3 · · ·Cash flow −c0 c1 c2 c3 . . .

Term Assurance. A premium c0 is paid at time 0. If death occurs before time n, then the beneficiary receivesan amount c. In this case, the time of any payout is uncertain. There are many variations—for example, thepolicy may be purchased by a sequence of annual premiums.

Loans and Mortgages. The borrower initially receives a loan and then repays it by a series of payments whichconsist of partial repayment of the loan capital and interest.

Motor and Other General Insurance. The insurance company receives a premium (positive cash flow) at thestart of the year. The amounts and times of any claims (or negative cash flows) are uncertain. The negativecash flows may occur after the end of the period of cover—especially for personal injury claims.



2 Net present value and discounted cash flow

2.1 Net present value. The net present value of an investment is the sum of all incoming payments (whichare regarded as positive) and all outgoings (which are regarded as negative) after converting all payments intotime 0 values. The cash amounts must be discounted to allow for the fact that cash received tomorrow is worthless than cash received today—the reason for the term discounted cash flow.

Example 2.1a. An investment of c0 is made for one time unit at a simple interest rate of i per time unit. Find thenet present value at time 0.

Solution. The cash-flow sequence can be represented by the following table:

Time 0 1Cash flow, a −c0 c1 = c0(1 + i)

The net present value at time 0 is

NPV(a) = −c0 +1

1 + ic1 = 0

In general, suppose c = (c0, c1, . . . , cn) denotes the cash-flow sequence of payment c0 at time 0 (the presenttime), and payment ck after of k years for k = 1, . . . , n. Suppose the current interest rate on investments is iper annum. Then the net present value is

NPV(c) = c0 +c1

1 + i+

c2

(1 + i)2 + · · · + cn(1 + i)n

=n∑j=0

cj(1 + i)j

(2.1a)

Equation (2.1a) is sometimes called the discounted cash flow or (DCF ) formula.Clearly NPV(c) decreases as i increases. Also, it decreases faster when later payments such as cn are larger.

If the net present value of a project is positive, then the project is considered profitable; conversely, a negativeNPV is evidence that a project should not proceed.

Example 2.1b. Consider the two cash flow sequences a = (12, 12, 12, 20, 24) and b = (20, 18, 14, 12, 12) attimes t = 0, . . . , 4. Find the net present values of the two cash-flows assuming an interest rate of (a) 3% and (b) 10%.Note that

∑aj = 80 and

∑bj = 76 and so a seems preferable.

Solution. (a) For 3%, we have NPV(a) = 12 + 12ν + 12ν2 + 20ν3 + 24ν4 = 74.59 where ν = 1/1.03. SimilarlyNPV(b) = 72.31 and so a is preferable to b.(b) For 10%, we have NPV(a) = 64.25 and NPV(b) = 65.15 and so b is preferable to a.If interest rates are high, b is preferable because the cash is received earlier—and in principle this cash could beinvested at this high interest rate.

Example 2.1c. Consider the following two investment projects:• ProjectA. Purchase a 4 year lease on an office block for £1,000,000 with returns of £300,000 at time 1, £325,000at time 2, £345,000 at time 3 and £360,000 at time 4.• Project B. Invest in a 4-year bond which pays a nominal 12% interest p.a. payable half-yearly, currently priced at£95 per £100 and redeemable at par.

Solution. For a £95 investment in a bond, the cash flows are as follows:Time in half-years 0 1 2 3 4 5 6 7 8Cash flow −95 6 6 6 6 6 6 6 106

Let ν = 1/(1 + r) denote the 6-month discount factor. Then−95 + 6ν + 6ν2 + 6ν3 + 6ν4 + 6ν5 + 6ν6 + 6ν7 + 106ν8 = 0

which leads to 101ν + 100ν8 − 106ν9 = 95 or 95(1 + r)9 − 101(1 + r)8 − 100r + 6 = 0. Solving numerically givesr = 0.06832, or an effective annual rate of interest of 1 + i = (1 + r)2 = 1.068322 = 1.141308 which is 14.13%.

Using this yield on the cash flows in project A we get the NPV:

−1,000,000 +300,000

1 + i+

325,000(1 + i)2 +

345,000(1 + i)3 +

360,000(1 + i)4 = 19,374.48

As NPV is positive, this suggests that the project A is preferable.However, the above analysis has not allowed for the differential risks of the two projects.

2 Cash Flows and Project Appraisal Sep 27, 2016(9:51) Section 2 Page 23

Example 2.1d. An investment of £50,000 is made in an annuity. The annuity pays out the amount £a at the end ofeach of the years 1, 2, . . . , 20. Suppose the investment gives a yield of 7%. Find a.

Solution. Let r = 0.07. Then

50,000 =a

1 + r+

a

(1 + r)2 + · · · + a

(1 + r)20 =a

r

[1− 1

(1 + r)20

]and so

a = 50,000r/[

1− 1(1 + r)20

]= 4,719.65 using r = 0.07.

Future cash flows and interest rates are usually uncertain. Hence it may be wise to calculate the NPV underseveral different assumptions.

2.2 General results. Consider the cash-flow sequences a = (a0, . . . , an) and b = (b0, . . . , bn). When is thecash-flow sequence a preferable to b? Here are two general results.

(a) If aj ≥ bj for every j = 0, . . . , n, then NPV(a) ≥ NPV(b) for all interest rates i ∈ [0,∞).

Proof: This is easy.

(b) Ifj∑k=0

ak ≥j∑k=0

bk for all j = 0, . . . , n

then NPV(a) ≥ NPV(b) for all interest rates i ∈ [0,∞).

Proof: (Not examinable.) The proof is by induction on n.First we prove for n = 0: the condition implies that a0 ≥ b0 and hence NPV(a) = a0 ≥ b0 = NPV(b) forall interest rates i.We now assume the result is true for n and we shall prove this implies the result is true for n + 1. Sosuppose a = (a1, . . . , an+1) and b = (b1, . . . , bn+1). Then

NPV(a)− NPV(b) =n+1∑k=0

ak − bk(1 + i)k

Consider separately the two cases: an+1 ≥ bn+1 and an+1 < bn+1. First suppose an+1 ≥ bn+1; then

NPV(a)− NPV(b) ≥n∑k=0

ak − bk(1 + i)k

and this is non-negative by the induction assumption. Second, suppose an+1 < bn+1; thenan+1 − bn+1

(1 + i)n+1 ≥an+1 − bn+1

(1 + i)n

and so

NPV(a)− NPV(b) =n−1∑k=0

ak − bk(1 + i)k

+an − bn(1 + i)n

+an+1 − bn+1

(1 + i)n+1

≥n−1∑k=0

ak − bk(1 + i)k

+an − bn + an+1 − bn+1

(1 + i)n=

n∑k=0

a′k − b′k(1 + i)k

where a′k = ak and b′k = bk for k = 0, 1, . . . , n− 1 and a′n = an + an+1 and b′n = bn + bn+1. Nowj∑k=0

a′k ≥j∑k=0

b′k for all j = 0, . . . , n and hencen∑k=0

a′k − b′k(1 + i)k

≥ 0 by the induction assumption.

Hence NPV(a)− NPV(b) ≥ 0.

More complicated results are possible, but they seem to have little utility.


2.3 Some standard examples.

Example 2.3a. Replacing a piece of capital equipment.Suppose an old machine has current value w0. For j = 1, 2, . . . , suppose the machine has value wj at the end ofyear j and has operating cost cj for year j. Typically the wj decrease with j due to depreciation, and the cj increasewith j.Suppose a new machine will cost W0 and will have value Wj at the end of year j of its operation. Also, its operatingcost for year j of its operation is Cj . Suppose the interest rate is i per annum.

By considering the sequence of cash flows over the first 5 years, decide whether the machine be replaced immediately,or at the end of year 1, or at the end of year 2, etc.

Solution. Suppose the machine is replaced at time 0 (the beginning of year 1). Then the sequence of cash flows is:a1 = (W0 + C1 − w0, C2, C3, C4, C5,−W5)

For replacement at start of year 2: a2 = (c1,W0 + C1 − w1, C2, C3, C4,−W4)For replacement at start of year 3: a3 = (c1, c2,W0 + C1 − w2, C2, C3,−W3)For replacement at start of year 4: a4 = (c1, c2, c3,W0 + C1 − w3, C2,−W2)

Clearly

NPV(a1) = W0 + C1 − w0 +C2

1 + i+

C3

(1 + i)2 +C4

(1 + i)3 +C5

(1 + i)4 −W5

(1 + i)5

Similarly for NPV(a2), etc. The best choice is the one with the smallest value for NPV.

Example 2.3b. Saving for a pension.Suppose someone plans to save the same amount £a at the beginning of every month for the next 240 months. Thisperson then intends to withdraw £c at the beginning of every month for the subsequent 360 months. Assume theinterest rate is i per annum compounded monthly. Find an expression for a in terms of i and c.

Solution. The monthly interest rate is i/12. Let α = 1 + i/12.At the end of 240 months, the capital accumulated is:

aα240 + aα239 + · · · + aα = aαα240 − 1α− 1

Using the same time point, the value of all the withdrawals is

c +c

α+ · · · + c

α359 = cα360 − 1

α359(α− 1)Setting these two quantities equal gives:

a = cα360 − 1

α360(α240 − 1)

Alternatively, calculations can be done in terms of the present value. At time 0, the value of the savings is:

a +a

α+ · · · + a

α239 = aα240 − 1

α239(α− 1)At time 0, the value of the withdrawals is:

c

α240 +c

α241 + · · · + c

α599 = cα360 − 1

α599(α− 1)Setting these quantities equal gives the same result as before.

Example 2.3c. Mortgage calculations. Suppose the capital borrowed is c0 and the interest rate is i per annumcompounded monthly. The capital is to be repaid by n equal payments of size x at the end of each month.(a) Express x in terms of n, c0 and i.(b) After the payment at the end of month j, what is the outstanding loan?(c) How much of the loan is reduced by the payment made at the end of month j?

Solution. (a) Let α = 1 + i/12. The present value of the n monthly payments is:x

α+x

α2 + · · · + x

αn= x

αn − 1αn(α− 1)

Setting this equal to c0 gives:

x = c0αn(α− 1)αn − 1

(b) The value at time 0 of the amount repaid by the end of month j is the present value of the first j payments:x

α+x

α2 + · · · + x

αj= x

αj − 1αj(α− 1)

=c0α

n−j(αj − 1)αn − 1


Hence the value at time 0 of the loan outstanding at the end of month j is

c0 −c0α

n−j(αj − 1)αn − 1

= c0αn−j − 1αn − 1

Hence rj , the amount of loan outstanding at the end of month j is

αj × c0αn−j − 1αn − 1

= c0αn − αj

αn − 1An alternative solution is obtained by using the following recurrence relations: r0 = c0 and rj+1 = αrj − x forj = 0, 1, . . . , n− 1.

(c) The reduction in the outstanding loan at the end of month j is

rj−1 − rj = c0αj−1(α− 1)αn − 1

A check of the validity of this last result can be given by adding over j; this gives the total loan repaid:n∑j=1

c0αj−1(α− 1)αn − 1

= c0

Note that the amount of loan repaid at the end of the first month is c0(α− 1)/(αn − 1) and this amount increases bythe factor α every month.This type of calculation is considered in depth in section 4 of chapter 3 on page 57.

2.4 The NPV of a discrete cash flow. Suppose ν(t) denotes the present value (time 0) of the amount 1 attime t for t ≥ 0. Hence ν(0) = 1.Suppose 0 ≤ t1 < t2 < · · · < tn and we have the cash-flow: c1 at time t1, c2 at time t2, . . . , and cn at time tn.

Time t1 t2 . . . tnCash flow, a c1 c2 . . . cn

The present value (at time 0) of the cash flow a is

NPV(a) =n∑j=1

cjν(tj)

2.5 The NPV of a continuous cash flow. Now suppose we have a function ρ : (0,∞)→ R which representsa continuous rate of payment. This means that if m(t) denotes the total payment received in [0, t], then

m′(t) = ρ(t)

It follows that m can be obtained from ρ by using the relation:

m(t) =∫ t

0ρ(u) du

Now the cash received in (t, t + h] is m(t + h) −m(t) = hρ(u) for some u ∈ [t, t + h]. For h small, this hasNPV approximately equal to hρ(t)ν(t). This motivates the following definition:

The net present value (time 0) of the continuous rate of payment ρ : (0, t)→ R is

NPV(ρ) =∫ t

0ρ(u)ν(u) du

where ν(t) denotes the present value (time 0) of the amount 1 at time t for t ≥ 0.

(2.5a)

2.6 The NPV of a cash flow with discrete and continuous payments. Now suppose we have the discretepayments of paragraph 2.4 and the continuous payment stream ρ : (0, t)→ R. To get the net present value, wejust need add the net present values for the discrete and continuous payments:

NPV(a, ρ) =n∑j=1

cjν(tj) +∫ t

0ρ(u)ν(u) du

If there are infinitely many discrete payments and ρ : (0,∞)→ R, then

NPV(a, ρ) =∞∑j=1

cjν(tj) +∫ ∞

0ρ(u)ν(u) du (2.6a)

provided the sum and integral converge—and they will converge for any sensible practical example.


If there are both incoming and outgoing payments, then the net present value is just the difference between thenet present values of the positive and negative cash flows.

For the special case of a constant interest rate i we have ν(t) = 1/(1 + i)t and so equation (2.6a) becomes

NPV(a, ρ) =∞∑j=1

cj(1 + i)tj

+∫ ∞

0

ρ(u)(1 + i)u

du

For A(tL), the accumulated value at time tL, just use:

A(tL) =NPV(a, ρ)ν(tL)

and in general we have

(value at time t1 of cash flow)× ν(t1) = (value at time t2 of cash flow)× ν(t2) = NPV

For example, if we have a constant interest rate i and the project ends at time tL, where tL ∈ [tn, tn+1) andρ(u) = 0 for u > tL, then A(tL), the accumulated value of the project at time tL is:

(1 + i)tLNPV(a, ρ) =n∑j=1

cj(1 + i)tL−tj +∫ tL

0ρ(u)(1 + i)tL−u du (2.6b)

2.7 The NPV in terms of the force of interest, δ. Now we know that ν(t) = exp(−∫ t

0 δ(u) du)

is the presentvalue of the amount 1 at time t, where δ is the corresponding force of interest function. Hence equation (2.6a)becomes:

NPV(a, ρ) =∞∑j=1

cj exp(−∫ tj

0δ(u) du

)+∫ ∞

0ρ(u) exp

(−∫ u

0δ(s) ds

)du

Now the accumulated value at time tL of the amount 1 at time t1 isν(t1)ν(tL)

= exp(∫ tL

t1

δ(u) du)

and this result holds if tL > t1 or t1 < tL. In general, for the cash flow (a, ρ), the accumulated value at time tLis

A(tL) =NPV(a, ρ)ν(tL)

=∞∑j=1

cj exp

(∫ tL

tj

δ(u) du

)+∫ ∞

0ρ(u) exp

(∫ tL

uδ(s) ds

)du

2.8 NPV of an interest stream. Differentiating the relation ν(t) = exp(−∫ t

0 δ(u) du)

shows that ν ′(t) =−δ(t)ν(t) for all t ≥ 0. Hence∫ t

0δ(u)ν(u) du = −

∫ t

0ν ′(u) du = 1− ν(t) because ν(0) = 1

Rearranging gives

ν(t) +∫ t

0δ(u)ν(u) du = 1 for all t ≥ 0. (2.8a)

Comparing the second term on the left hand side with equation (2.5a) shows that equation (2.8a) can beinterpreted as follows: the net present value of the amount 1 at time t plus the net present value of the intereststream δ over the interval (0, t) is equal to 1. Hence equation (2.8a) represents the situation where we have aconstant amount of 1 invested and the interest is continuously withdrawn. The term interest stream means acontinuous payment stream.

Provided ν(t)→ 0 as t→∞, equation (2.8a) implies∫ ∞0

δ(u)ν(u) du = 1

This says that the net present value of the interest stream δ : (0,∞)→ R is 1.

2 Cash Flows and Project Appraisal Sep 27, 2016(9:51) Exercises 3 Page 27

2.9

Summary.• The net present value (time 0) of the continuous rate of payment ρ : (0, t)→ R is

NPV(ρ) =∫ t

0ρ(u)ν(u) du =

∫ t

0ρ(u) exp

(−∫ u

0δ(s) ds

)du (2.9a)

where ν(t) denotes the present value (time 0) of the amount 1 at time t for t ≥ 0.For a constant interest rate i, equation (2.9a) becomes

NPV(ρ) =∫ t

0

ρ(u)(1 + i)u

du (2.9b)

• The accumulated value at time t of the continuous rate of payment ρ : (0, t)→ R is

A(t) =∫ t

0ρ(u)

ν(u)ν(t)

du =∫ t

0ρ(u) exp

(∫ t

uδ(s) ds

)du (2.9c)

For a constant interest rate i, equation (2.9c) becomes

A(t) =∫ t

0ρ(u)(1 + i)t−u du (2.9d)


1. Consider the cash-flows a = (100, 2500) and b = (2500, 10) at times 0 and 1.(a) Suppose the interest rate is 1%. Show that NPV(a) > NPV(b).(b) Suppose the interest rate is 10%. Show that NPV(a) < NPV(b).(c) Suppose the interest rate is 1%. Show that the cash flow a can be transformed by borrowing and investing at

rate 1% into a cash-flow c = (c0, c1) with c0 ≥ b0 and c1 ≥ b1.(d) Suppose the interest rate is 10%. Show that the cash flow b can be transformed by borrowing and investing at

rate 1% into a cash-flow c = (c0, c1) with c0 ≥ a0 and c1 ≥ a1.(e) In general, suppose a = (a0, . . . , an) and b = (b0, . . . , bn) are 2 cash-flows with NPV(a) ≥ NPV(b) at interest

rate i per annum. Show that the cash-flow sequence a can be transformed by investing and borrowing at rate iinto a cash-flow sequence c = (c0, . . . , cn) with c0 ≥ b0, . . . , cn ≥ bn.

2. Suppose a fund has value A(0) at time 0. Suppose further that investments are deposited into a fund at a continuousrate with value ρ(t) at time t and the force of interest function has value δ(t) at time t.(a) Find A(t), the value of the fund at time t.(b) Suppose the force of interest is constant and corresponds to the effective annual interest rate i. Express A(t) in

terms of i and ρ.

3. An individual makes an investment of £4m per annum in the first year, £6m per annum in the second year and £8mper annum in the third year. The investments are made continuously throughout each year. Calculate the accumulatedvalue of the investments at the end of the third year at a rate of interest of 4% per annum effective.


4. The force of interest δ(t) at time t is given by

δ(t) =

0.06 if 0 < t ≤ 6;0.05 + 0.0002t2 if 6 < t ≤ 12.

Calculate the accumulated value at time t = 12 of a continuous payment stream of £100 per annum payable fromtime t = 0 to time t = 6. (Institute/Faculty of Actuaries Examinations, April 2000) [5]

5. A particular floating note pays interest linked to an index of local floating interest rates. The note was purchased fora price of 99 on 1 January 2000 and sold for a price of 101 on 31 December 2000. The nominal value of the noteis 100 and the note paid interest of 6.5% of the nominal value on 30 June 2000 and 6.6% of the nominal value on31 December 2000. Calculate the nominal rate of return per annum convertible half-yearly earned by the investor.

(Institute/Faculty of Actuaries Examinations, April, 2002) [5]

6. The force of interest, δ(t), is a function of time and at any time t (measured in years) is given by:

δ(t) =

0.05 if 0 < t < 80.04 + 0.0004t2 if 8 ≤ t ≤ 15

Calculate the accumulated value at time t = 15 of a continuous payment stream of £50 per annum payable from timet = 0 to t = 8. (Institute/Faculty of Actuaries Examinations, April 2004) [6]


7. The force of interest δ(t) is a function of time and at any time, measured in years, is given by the formula

δ(t) =

0.04 + 0.01t if 0 ≤ t ≤ 4;0.12− 0.01t if 4 < t ≤ 8;0.06 if 8 < t.

Calculate the present value at time t = 0 of a payment stream, paid continuously from time t = 9 to t = 12, underwhich the rate of payment at time t is 50e0.01t. (Institute/Faculty of Actuaries Examinations, April 2007) [6]

8. A businessman is considering an investment which requires an outlay of £60,000 and a further outlay of £25,000 ineight months’ time.Starting 2 years after the initial outlay, it is estimated that income will be received continuously for 4 years at a rateof £5,000 per annum, increasing to £9,000 per annum for the next 4 years, then increasing to £13,000 per annum forthe following 4 years and so on, increasing by £4,000 per annum every 4 years until the payment stream stops afterincome has been received for 20 years (i.e. 22 years after the initial outlay). At the point when the income ceases,the investment can be sold for £50,000.Calculate the net present value of the project at a rate of interest of 9% per annum effective.



δ(t) = 0.07− 0.005t for t ≤ 8

0.06 for t > 8(a) Calculate the accumulation at time t = 10 of £500 invested at time t = 0.(b) Calculate the present value at time t = 0 of a continuous payment stream at the rate of £200e0.1t paid from

t = 10 to t = 18. (Institute/Faculty of Actuaries Examinations, April 2005) [3+5=8]

10. A university student receives a 3-year sponsorship grant. The payments under the grant are as follows:Year 1 £5,000 per annum paid continuouslyYear 2 £5,000 per annum paid monthly in advanceYear 3 £5,000 per annum paid half-yearly in advance

Calculate the total present value of these payments at the beginning of the first year using a rate of interest of 8% perannum convertible quarterly. (Institute/Faculty of Actuaries Examinations, April 2005) [3+5=8]

11. The force of interest δ(t) is a function of time and at any time t, measured in years, is given by the formula:

δ(t) =

0.04 if 0 < t ≤ 5;0.008t if 5 < t ≤ 10;0.005t + 0.0003t2 if 10 < t.

(i) Calculate the present value of a unit sum of money due at time t = 12.(ii) Calculate the effective annual rate of interest over the 12 years.

(iii) Calculate the present value at time t = 0 of a continuous payment stream that is paid at the rate of e−0.05t perunit time between time t = 2 and time t = 5.

(Institute/Faculty of Actuaries Examinations, April 2006) [5+2+3=10]

12. The force of interest δ(t) at time t is at + bt2 where a and b are constants. An amount of £100 invested at time t = 0accumulates to £150 at time t = 5 and £230 at time t = 10.

(i) Calculate the values of a and b.(ii) Calculate the constant force of interest that would give rise to the same accumulation from time t = 0 to time

t = 10.(iii) At the force of interest calculated in (ii), calculate the present value of a continuous payment stream of 20e0.05t

paid from time t = 0 to time t = 10. (Institute/Faculty of Actuaries Examinations, September 2006) [5+2+4=11]

13. The force of interest δ(t) is:

δ(t) =

0.04 for 0 < t ≤ 5;0.01(t2 − t) for 5 < t.

(i) Calculate the present value of a unit sum of money due at time t = 10.(ii) Calculate the effective rate of interest over the period t = 9 to t = 10.

(iii) (a) In terms of t, determine an expression for ν(t), the present value of a unit sum of money due during theperiod 0 < t ≤ 5.

(b) Calculate the present value of a payment stream paid continuously for the period 0 < t ≤ 5, where the rateof payment, ρ(t), at time t is e0.04t.



14. The force of interest δ(t) is a function of time and at any time, measured in years, is given by the formula

δ(t) = 0.03 + 0.01t 0 ≤ t ≤ 8

0.05 8 < t(i) Derive, and simplify as far as possible, expressions for ν(t) where ν(t) is the present value of a unit sum of

money due at time t.(ii) (a) Calculate the present value of £500 due at the end of 15 years.

(b) Calculate the rate of discount per annum convertible quarterly implied by the transaction in (a).(iii) A continuous payment stream is received at rate 10e−0.02t units per annum between t = 10 and t = 14. Calculate

the present value of the payment stream.(Institute/Faculty of Actuaries Examinations, September 2002) [5+4+4=13]


δ(t) = 0.04 + 0.01t for 0 ≤ t ≤ 10

0.05 for t > 10(i) Derive and simplify as far as possible, expressions for ν(t) where ν(t) is the present value of a unit sum of

money due at time t.(ii) (a) Calculate the present value of £1,000 due at the end of 15 years.

(b) Calculate the annual effective rate of discount implied by the transaction in (a).(iii) A continuous payment stream is received at the rate of 20e−0.01t units per annum between t = 10 and t = 15.

Calculate the present value of the payment stream.(Institute/Faculty of Actuaries Examinations, September 2007) [5+4+4=13]

16. The force of interest per unit time at time t, δ(t), is given by:

δ(t) =

0.1− 0.005t for t < 6;0.07 for t ≥ 6.

(i) Calculate the total accumulation at time 10 of an investment of £100 made at time 0 and a further investment of£50 made a time 7.

(ii) Calculate the present value at time 0 of a continuous payment stream at the rate £50e0.05t per unit time receivedbetween time 12 and time 15. (Institute/Faculty of Actuaries Examinations, April 2013) [4+5=9]

17. The force of interest, δ(t), is a function of time and at any time t, measured in years, is given by the formula:

δ(t) =

0.06 0 ≤ t ≤ 40.10− 0.01t 4 < t ≤ 70.01t− 0.04 7 < t

(i) Calculate the value at time t = 5 of £1,000 due for payment at time t = 10.(ii) Calculate the constant rate of interest per annum convertible monthly which leads to the same result as in (i)

being obtained.(iii) Calculate the accumulated amount at time t = 12 of a payment stream, paid continuously from time t = 0 to

t = 4, under which the rate of payment at time t is ρ(t) = 100e0.02t.(Institute/Faculty of Actuaries Examinations, April, 2008) [5+2+6=13]

18. The force of interest, δ(t), at time t is given by:

δ(t) =

0.04 + 0.003t2 for 0 < t ≤ 5;0.01 + 0.03t for 5 < t ≤ 8;0.02 for t > 8.

(i) Calculate the present value (at time t = 0) of an investment of £1,000 due at time t = 10.(ii) Calculate the constant rate of discount per annum convertible quarterly, which would lead to the same present

value as that n part (i) being obtained.(iii) Calculate the present value (at time t = 0) of a continuous payment stream payable at the rate of 100e0.01t from

time t = 10 to t = 18. (Institute/Faculty of Actuaries Examinations, April 2012) [4+2+4=10]

19. The force of interest, δ(t), is a function of time and at any time t, measured in years, is given by the formula

δ(t) = 0.03 + 0.01t for 0 ≤ t ≤ 9;

0.06 for 9 < t.(i) Derive, and simplify as far as possible, expressions for ν(t) where ν(t) is the present value of a unit sum of

money due at time t.(ii) (a) Calculate the present value of £5,000 due at the end of 15 years.

(b) Calculate the constant force of interest implied by the transaction in part (a).A continuous payment stream is received at rate 100e−0.02t units per annum between t = 11 and t = 15.(iii) Calculate the present value of the payment stream.



20. An individual can obtain a force of interest per annum at time t, measured in years, as given by the formula:

δ(t) = 0.03 + 0.005t for 0 ≤ t ≤ 3;

0.005 for t > 3.(i) Determine the amount the individual would need to invest at time t = 0 in order to receive a continuous payment

stream of £5,000 per annum from time t = 3 to time t = 6.

(ii) Determine the equivalent constant annual effective rate of interest earned by the individual in part (i).

(iii) Determine the amount an individual would accumulate from the investment of £300 from time t = 0 to timet = 50. (Institute/Faculty of Actuaries Examinations, September 2015) [5+3+2=10]

4 Equations of value and the internal rate of return

4.1 The equation of value and the yield or IRR. Suppose an investment of £100 grows to £125 after oneyear. Then the rate of return is 0.25. Similarly, if the principal c0 grows to c1 after one year, then the rate ofreturn r is equal to

r =c1

c0− 1

Alternatively, it is the solution of

c0 =c1

1 + r

If the prevailing rate of interest is lower than the internal rate of return of a project, then that suggests theproject is profitable.

More generally, consider the following cash flow a where the interest rate is i and 1 + i = eδ:

Time t1 t2 · · · tnCash flow, a c1 c2 · · · cn

Then the following equation for δ:

NPV(a) =n∑j=1

cje−δtj = 0

is called the equation of value for δ for the cash flow c. The equation can also be written as an equation for i:

NPV(a) =n∑j=1

cj(1 + i)tj

= 0

and is then also called the yield equation for i or the equation of value for i.

Suppose we also have the net continuous payment stream ρ. Then the equation of value for δ is:

NPV(a, ρ) =n∑j=1

cje−δtj +

∫ ∞0

ρ(t)e−δt dt = 0

and the yield equation or equation of value for i is

NPV(a, ρ) =n∑j=1

cj(1 + i)tj

+∫ ∞

0

ρ(t)(1 + i)t

dt = 0

Define f : [0,∞)→ R by

f (i) =n∑j=1

cj(1 + i)tj

+∫ ∞

0

ρ(t)(1 + i)t

dt

In general, the equation f (i) = 0, where f (i) = NPV(a, ρ) is considered as a function of i, is called the equationof value for a project. If the equation f (i) = 0 has exactly one root i0 with i0 > −1 then the yield per unit time(or internal rate of return or money weighted rate of return) is defined to be i0.


4.2 Further results on the IRR. Consider the following setup: we have the sequence of cash flows: a =(−c0, b1, . . . , bn). This corresponds to an initial investment of c0 which leads to repayments b1, . . . , bn in thefollowing n years. The internal rate of return, r∗, of the cash-flow sequence a = (−c0, b1, . . . , bn) is then thesolution for r of

c0 =n∑j=1

bj(1 + r)j

The following result shows that there is a unique solution for the IRR for this situation.

Suppose c0 > 0, bj ≥ 0 for j = 1, . . . , n− 1 and bn > 0. Let

g(r) =n∑j=1

bj(1 + r)j

for r ∈ (−1,∞)

Thenlimr→−1

g(r) =∞, limr→∞

g(r) = 0, and g(r) ↓↓ as r ↑↑ for r ∈ (−1,∞).

It follows that there is a unique r∗ ∈ (−1,∞) with g(r∗) = c0. Using g(0) =∑n

j=1 bj gives the result that

r∗ > 0 ifn∑j=1

bj > c0, r∗ < 0 ifn∑j=1

bj < c0 and r∗ = 0 ifn∑j=1

bj = c0

The last result is just common sense: the rate of return is positive iff the total amount returned exceeds theinitial investment.If the interest rate equals the internal rate of return r∗, then NPV(a) = −c0 + g(r∗) = 0. Now suppose theinterest rate is i. Then NPV(a) > 0 iff i < r∗ and NPV(a) < 0 iff i > r∗. In words, if the interest rate isgreater than the internal rate of return, then the net present value is negative.The above result can be generalised as follows (see exercise 3 on page 35):

Suppose we have the cash flow a = (−c0,−c1, . . . ,−cm, b0, b1, . . . , bn) at times t0 < t1 < · · · < tm < t′0 <t′1 < . . . < t′n where all cj > 0 and all bj > 0. Then there is a unique solution for the IRR.

Example 4.2a. An investment of £100 returns £70 at the end of year 1 and £70 at the end of year 2. What is theinternal rate of return, r∗?

Solution. The sequence of cash-flows is a = (−100, 70, 70). So r∗ satisfies

−100 +70

1 + r∗+

70(1 + r∗)2 = 0

Let x = 1/(1+r∗). Hence 70x2 +70x−100 = 0. We need the root with x > 0 and so x = 0.796 and r∗ = (1/x)−1 =0.257. The internal rate of return is 25.7%.

Three payments leads to a cubic, and so on. The solution for r∗ will often have to be obtained numerically, butthis is usually easy as g(r) is monotonic.

4.3 Numerically finding the IRR. Starting values for a numerical search are often required. Here are fourpossibilities:• (a) Approximate 1/(1 + r)j = (1 + r)−j by 1− jr. Then the equation

c0 =c1

1 + r+

c2

(1 + r)2 + · · · + cn(1 + r)n

gives the following approximation r0 for r∗:

c0 =n∑j=1

cj − r0

n∑j=1

jcj or r0 =

∑nj=1 cj − c0∑nj=1 jcj

• (b) Convert to c0(1 + r)n = c1(1 + r)n−1 + c2(1 + r)n−2 + · · · + cn. Then approximate (1 + r)j by 1 + jr.

• (c) Assume all future cash-flows equal the average cash-flow. This means that we approximateTime 0 1 2 . . . n− 1 nCash flow −c0 c1 c2 . . . cn−1 cn

by


Time 0 1 2 . . . n− 1 nCash flow −c0 x x . . . x x

where x =∑n

j=1 ci

/n. This leads to

c0 =x

r

[1− 1

(1 + r)n

]= xan,r where an,r =

1r

[1− 1

(1 + r)n

]and values of an,r can be found in actuarial tables. There is further discussion of an,r in the next chapter.

• (d) Suppose the cash flow has the following form:

Time 0 1 2 . . . n− 1 nCash flow −c0 x x . . . x cn

There is an interest payment of x for each of the n years plus a final payment of cn−x. Hence the capital gainis cn − x − c0 and this capital gain can be approximated by a payment of (cn − x − c0)/n every year. So theapproximate rate of return is

i =x + (cn − x− c0)/n

c0

If there is a capital gain, we will have x/c0 < i < [x + (cn − x − c0)/n]/c0; if there is a capital loss, we willhave x/c0 > i > [x + (cn − x− c0)/n]/c0. This will usually be the best method for cash flows of this form.

Example 4.3a. (a) Consider the following cash-flow sequence

Time 0 1 2 3 4 5 6 7 8 9 10Cash flow −50 4 4 4 4 4 4 4 4 4 59

Use method (d) described above to find an initial approximation to the internal rate of return, r∗.

Solution. The return per year is 4 + (55− 50)/10 = 4.5. Hence r∗ ≈ 4.5/50 = 0.09. In fact, r∗ = 0.0866868.

Example 4.3b. (a) Consider the cash-flow sequence a = (−87, 25,−40, 60, 60) at times 0, 1, 2, 3 and 4. Use thefirst 3 methods described above to find initial approximations to the internal rate of return, r∗.(b) Use the first 3 methods to find the yield on project B in example (2.1c) on page 22.

Solution. (a) The first method gives r0 =∑cj/∑jcj = 18/365 = 0.049. The second method gives r0 =

18/293 = 0.061. For the third method, we have x = 105/4 = 26.25. So we want r with a4 ,r = 87/26.25 = 3.312.From actuarial tables, a4 ,0.07 = 3.387 and a4 ,0.08 = 3.312; so take r = 0.075. The actual value of r∗ is 0.056.(b) The first method gives r0 =

∑cj/∑jcj = 53/1016 = 0.052. The second method gives 53/592 = 0.0895.

For the third method, we have x = 148/8 = 18.5. So we want r with a8 ,r = 95/18.5 = 190/37 = 5.135. Fromactuarial tables, a8 ,0.11 = 5.146 and a8 ,0.115 = 5.0455, so take r = 0.11.The actual value is r = 0.06832. The third method gives a poor approximation because the mean of 18.5 is a poorsummary of the sequence (6, 6, 6, 6, 6, 6, 6, 106).

4.4 Comparison with NPV. If the interest rate is i, then NPV(i) is the present value of the cash-flows of theproject. The project is profitable if NPV(i) > 0.Suppose r∗, the yield or IRR exists, and NPV(i) > 0 for i < r∗ and NPV(i) < 0 for i > r∗. Then the projectis profitable if and only if i < r∗.

4.5 Problems with the use of the IRR. Consider the cash-flow sequence a = (c0, c1, . . . , cn). Let x∗ =1/(1 + r∗). Then x∗ is the solution for x of

c0 + c1x + c2x2 + · · · + cnxn = 0

This equation has n roots for x. Any root for x which satisfies x > 0 leads to a solution for r with r > −1.There may be more than one root with r > −1. Hence the internal rate of return cannot be defined.Even if there is a unique root, it is possible that g(r) is not monotone. It then follows that the property thatNPV is positive for interest rates i on one side of r∗ and negative for interest rates on the other side of r∗ maynot hold. So again it is not sensible to define the IRR.


Example 4.5a. Consider the following sequence of cash-flows:

Time 0 1 2 3Cash flow −32 128 −166 70

Find the internal rate of return, r∗.

Solution. The equation c0 + c1x + c2x2 + · · · + cnxn = 0 gives −32 + 128x − 166x2 + 70x3 = 0. We can easily see

that x = 1 is a solution. Hence we have 0 = (x − 1)(70x2 − 96x + 32) = (x − 1)(7x − 4)(10x − 8). Hence x = 1,4/7, or 4/5. Hence r∗ = 0, 0.75, or 0.25. There is no unique value for the IRR. Also NPV(r) < 0 for r ∈ (0, 0.25)and NPV(r) > 0 for r ∈ (0.25, 0.75).

Example 4.5b. Find the IRR for the following cash-flow.

Time 0 1 2Cash flow 2 −4 3

Solution. We have 2− 4x + 3x2 = 0 and this has no real roots.

The main disadvantages of using the IRR are:

• Some cash-flow sequences do not have an IRR: there may be no or multiple solutions for r∗.• The IRR measure does not use the value of the currently available interest rates. It is solely an internalmeasure—that is why it is called the internal rate of return. In particular, if an investment accumulates fundsfor a final payment, then the IRR calculation assumes that the sinking fund earns interest at the internal rate ofreturn. Similarly, if a loan is required to finance a project, then the IRR calculation assumes the interest rate onthe loan is at the internal rate of return.• NPV and IRR may lead to different conclusions—an example is given in the next section.

The NPV method is generally preferred.

5 Comparing two projects5.1 A worked example.

Example 5.1a. Consider the following two cash-flow sequences.

Time 0 1 2 3Project A Cash flow, a −80 96 1 5Project B Cash flow, b −80 10 10 90

(a) Find r∗A, the IRR for project A and r∗B , the IRR for project B and show that r∗A > r∗B . This suggests that projectAis preferable to project B.(b) Show that NPV(b) > NPV(a) if the interest rate is r = 0.04.(c) Treating NPVi(a) and NPVi(b) as functions of the interest rate i, plot their graphs for i ∈ (0, 1).

Solution. (a) For project A, solve −80 + 96x + x2 + 5x3 = 0. This gives (20 + x + x2)(−4 + 5x) = 0. Hence x∗ = 4/5and r∗A = 1/x∗ − 1 = 1/4.For project B, solve −80 + 10x + 10x2 + 90x3 = 0. This gives 10(1 + x + x2)(−8 + 9x) = 0. Hence x∗ = 8/9 andr∗B = 1/x∗ − 1 = 1/8.(b) If r = 0.04 = 1/25 then NPV(a) = 17.677 and NPV(b) = 18.87 and project B is preferable to project A!!(c) This is shown in figure (5.1a). The value of i where the two graphs cross is called the cross-over rate.

The previous example shows that the choice of investment depends on the rate of interest at which the investorcan borrow or invest money. The sum of the returns is 102 for project A and 110 for project B. However,the returns from project A are obtained quickly. Hence, if interest rates are high, then project A is preferablebecause these early returns can be invested.

More complex graphs than figure (5.1a) are possible—it is possible to have more than one cross-over point.

5.2 Different interest rates for lending and borrowing. Usually, an investor has to pay a higher rateof interest on borrowings (iB) than he obtains on investments (iI ). The size of the differential depends onhis credit worthiness, size of the amounts, time of loan, etc. The concepts of NPV and IRR are then notmeaningful.


interest rate

NPV

0.0 0.05 0.10 0.15 0.20 0.25 0.30

-30

-20

-10

0

10

20

30

NPVi(b)

NPVi(a)

Figure 5.1a. Plot of NPVi(a) and NPVi(b) as function of i for i ∈ (0, 0.3)(wmf/npv1,250pt,175pt)

5.3 Discounted payback period. For many projects, there is an initial cash outflow followed by a sequenceof cash inflows. Let A(t) denote the accumulated value of the investment at time t; then

A(t) =∑tj≤t

ctj (1 + iB)t−tj +∫ t

0ρ(s)(1 + iB)t−s ds

Typically A(t) will be initially negative and then become positive. Let t1 denote the smallest value of t withA(t) ≥ 0. This time t1 is called the discounted payback period or DPP—it is the smallest time such that theaccumulated value of the project becomes positive. Equivalently, the DPP is the smallest time t such that theNPV of payments up to time t is positive. The value of the DPP depends on iB but not on iI .

If iB = 0 then the quantity corresponding to t1 is called the payback period. Thus the payback period is thetime when the net cash flow (incomings less outgoings) is positive. It is similar to the DPP—but it ignoresdiscounting. This means that it ignores the effect of interest. It is therefore an inferior measure and shouldrarely be used.

Suppose the project ends at time k.• If A(k) < 0 (or equivalently NPViB (a, ρ) < 0) then the project is not profitable and the DPP does not exist.• If the project is profitable and DPP = t1 then the accumulated profit when the project terminates at time k is

A(t1)(1 + iI )k−t1 +∑

k≥tj>t1

ctj (1 + iI )k−tj +∫ k

t1

ρ(s)(1 + iI )t−s ds

Example 5.3a. Consider the following cash flow:

Time 0 1 · · · n− 1 nCash flow −C x · · · x x

This corresponds to an investment of C which leads to annual payments of x for n years.(a) Find an expression for the discounted payback period, t1.(b) If t1 ≤ n, find an expression for the accumulated profit at time n in terms of t1.Assume an annual interest rate of i1.

Solution. Now A(0) = −C and A(1) = C(1 + i1) + x. AlsoA(t) = −C(1 + i1)t + x

[(1 + i1)t−1 + · · · + (1 + i1) + 1

]for t ∈ 1, 2, . . . , n.

For t ≥ n + 1A(t) = −C(1 + i1)t + x

[(1 + i1)t−1 + · · · + (1 + i1)t−n+1 + (1 + i1)t−n

]Then t1 is the smallest integer t with A(t) ≥ 0 provided such a t1 exists.(b) If t1 ≤ n, then the accumulated profit at time n is

A(t1)(1 + i1)n−t1 + x[(1 + i1)n−t1−1 + · · · + (1 + i1) + 1

]


The DPP and payback period are measures of the time it takes for a project to become profitable. Howeverthey do not show how large or small the profit is. The DPP is especially useful if capital is scarce. A projectwhich has a smaller DPP than another may be regarded as preferable because the capital is released earlier andavailable for use elsewhere. However, the other project which has a longer DPP may have a higher NPV! Thiscan occur if the other project has a late large cash inflow.If the interest rate assumed in the calculation of the NPV is correct, then the project with the NPV is moreprofitable. It may be sensible to use a higher interest rate for discounting projects which are “longer” in orderto allow for the uncertainty in forecasting future interest rates. This leads on to the idea of sensitivity analysis.Because some of the quantities used in the calculation of the NPV are estimates, it is sensible to vary thesequantities one at a time to see how the value of the NPV varies with the assumptions.

5.4

Summary.• Equation of value: NPV(a, ρ) = 0 considered as an equation for i.• Internal Rate of Return (IRR): disadvantages. Numerical solution: methods for finding a startingvalue. Linear interpolation.• Discounted Payback Period (DPP). Payback period.


1. (Not examined) Write an R function which takes two arguments: vec.cashflow which represents the vector of cashflows, and r which represents a vector of interest rates. The function should return a matrix with two columns: thefirst column called interest and the second column called NPV.Thus the callirr( c(-87,25,-40,60,60), seq(0.01,0.3,by=0.001) )

should return a matrix like this:interest NPV

[1,] 0.010 14.434862816[2,] 0.011 14.087489548

. . .many lines omitted[46,] 0.055 0.288410285[47,] 0.056 0.005780673[48,] 0.057 -0.275585819

. . .many lines omitted[290,] 0.299 -43.014001386[291,] 0.300 -43.120233885

This shows the required IRR is 0.056.

2. Compare the following two investment projects:• Project A: Investment costing £50,000 with returns £8,000 each half-year for the next 4 years.• Project B: Government bonds which provide a yield of 7% every 6 months.

3. Suppose we have the cash flow a = (−c0,−c1, . . . ,−cm, b0, b1, . . . , bn) at times t0 < t1 < · · · < tm < t′0 < t′1 <. . . < t′n where all cj > 0 and all bj > 0. Prove there is a unique solution for the IRR.

4. An investor is considering two investment projects, A and B. Both involve outlays of £1 million. Project A willprovide a single incoming cash payment after 8 years of £1.7 million. ProjectB will provide incoming cash paymentsof £1 million after 8 years, £0.321 million after 9 years, £0.229 million after 10 years and £0.245 million after11 years.

(i) Determine the rate of interest (i′) at which the net present value of the two projects will be equal.(ii) By general reasoning or by illustrative calculation, show that at a positive rate of interest i∗ where i∗ < i′,

project B will have a higher net present value than project A.(Institute/Faculty of Actuaries Examinations, September 2004) [6]

5. A small technology company set up a new venture on 1 January 2001. The initial investment on that date was £2million with a further £1.5 million required on 1 August 2001.It is expected that on 1 January 2002, net income (i.e. income less running costs) will begin at the rate of £0.3million per annum and that the rate will increase by £0.1million per annum on 1 January of each subsequent year. It


is assumed that the net income will be received continuously throughout the project.The company expects to sell the business on 31 December 2011 for £3 million.Calculate the net present value of the venture on 1 January 2001 at a rate of interest of 6% per annum, convertiblehalf-yearly. (Institute/Faculty of Actuaries Examinations, April 2001) [8]

6. A computer manufacturer is to develop a new chip to be produced from 1 January 2008 until 31 December 2020.Development begins on 1 January 2006. The cost of development comprises £9 million payable on 1 January 2006and £12 million payable continuously during 2007.From 1 January 2008 the chip will be ready for production and it is assumed that income will be received half yearlyin arrear at a rate of £5 million per annum.

(i) Calculate the discounted payback period at an effective rate of interest of 9% per annum.(ii) Without doing any further calculations, explain whether the discounted payback period would be greater than,

less than or equal to that given in part (i) if the effective interest rate were substantially greater than 9% perannum. (Institute/Faculty of Actuaries Examinations, April 2005) [6+2=8]

7. A company has agreed to build and operate a toll bridge for a regional government. The company will invest £10million per annum for the first 2 years of the project, the investment being made continuously during this period.The bridge will then come into operation and the company will start to receive payments at the end of each year, thefirst payment occurring at the end of year 3 of the project. The amount of payment at the end of year 3 will be £8million, reducing by £0.5 million in each of the subsequent years until the annual amount is £3 million, after whichthe annual reduction will be £1 million. When the payments have reduced to zero, the company’s involvement in theproject will end.

(i) Calculate the net present value of the project at a rate of interest of 10% per annum effective.(ii) Explain whether the internal rate of return achieved on this project will be greater or less than 10% per annum

effective. (Institute/Faculty of Actuaries Examinations, September 2002) [7+2=9]

8. A motor manufacturer is to develop a new car model to be produced from 1 January 2002 for 6 years until 31 De-cember 2007. The development cost will be £33 million, of which £18 million will be incurred on 1 January 2000,£10 million on 1 July 2000 and £5 million on 1 January 2001.The production cost of each car is assumed to be incurred at the beginning of the calendar year of production andwill be £9,000 during 2002. The sale price of each car is assumed to be received at the end of the calendar year ofproduction. Both the production costs and the sale prices are assumed to increase by 5% on each 1 January, the firstincrease occurring on 1 January 2003. It is also assumed that 5,000 cars will be produced each year and that all willbe sold.The sale price of each car produced in 2002 is £12,100.

(i) Calculate the discounted payback period at an effective rate of interest of 9% per annum.(ii) Without doing any further calculations, explain whether the discounted payback period would be greater than,

equal to, or less than the period calculated in (i) if the effective rate of interest were substantially less than 9%per annum. (Institute/Faculty of Actuaries Examinations, April 2000) [9+3=12]

9. An investment project gives rise to the following cash flows. At the beginning of each of the first three years,£180,000 will be invested in the project. From the beginning of the first year until the end of the the twenty-fifthyear, net revenue will be received continuously. The initial rate of payment of net revenue will begin at £25,000 perannum. The rate of payment is assumed to grow continuously at a rate of 6% per annum effective.

(i) Calculate the net present value of the project at an effective rate of interest of 7% per annum.(ii) Calculate the discounted payback period of the project at an effective rate of interest of 7% per annum.

(iii) Calculate the annual effective rate of growth of net revenue which would be required if the project is to have azero net present value at an effective rate of interest of 7% per annum.


10. (i) An investor is deciding to invest in a project. Explain why the discounted payback period is a poorer decisioncriterion than net present value assuming the investor is not short of capital.An investor is considering two projects A and B. Project A involves the investment of £1 million at the outset.The only income to be received will be a payment of £3.5 million after ten years. Project B also involves theinvestment of £1 million at the outset. Income will be received from this project continuously. In the first yearthe rate of payment will be £0.08 million, in the second year £0.09 million, in the third year £0.10 million, withthe rate increasing by £0.01 million each year thereafter until the tenth year, after the end of which no furtherincome will be received.

(ii) Calculate the net present value of both investment projects at a rate of interest of 4% per annum effective.(iii) Show that the discounted payback period of project A is after that of project B (no further calculation is neces-

sary).(iv) In the light of your answer to (i) above, explain which project is the more desirable to an investor with unlimited

capital, and why. (Institute/Faculty of Actuaries Examinations, April 2002) [2+10+3+2=17]


11. A car manufacturer is to develop a new model to be produced from 1 January 2016 for six years until 31 Decem-ber 2021. The development costs will be £19 million on 1 January 2014, £9 million on 1 July 2014 and £5 millionon 1 January 2015.It is assumed that 6,000 cars will be produced each year from 2016 onwards and that all will be sold.

The production cost per car will be £9,500 during 2016 and will increase by 4% each year with the first increaseoccurring in 2017. All production costs are assumed to be incurred at the beginning of each calendar year.

The sale price of each car will be £12,600 during 2016 and will also increase by 4% each year with the first increaseoccurring in 2017. All revenue from sales is assumed to be received at the end of each calendar year.

(i) Calculate the discounted payback period at an effective rate of interest of 9% per annum.

(ii) Without doing any further calculations, explain whether the discounted payback period would be greater than,equal to, or less than the period calculated in part (i) if the effective rate of interest were substantially less than9% per annum. (Institute/Faculty of Actuaries Examinations, April 2013) [9+2=11]

12. (i) In respect of an investment project, define(a) the discounted payback period(b) the payback period

(ii) Discuss why both the discounted payback period and the payback period are inferior measures compared withthe net present value for determining whether to proceed with an investment project.

(iii) A consortium of investors is considering bidding to host a major athletics event. The project will be regarded asviable if it provides a positive net present value at a rate of interest of 10% per annum effective. The consortiumestimates the following cash flows will be generated by the event (all figures in £100 million):

CostsInitial costs of building To be incurred continuously at a rate of 1 per annum for five years starting

on 1 January 2006Running costs of the event To be incurred continuously at a rate of 1 per annum for 3 months starting

on 1 January 2011.Cost of making bid 0.2 to be incurred on 1 January 2004.

RevenueSale of television rights To be received continuously at a rate of 0.3 per annum for 3 months

starting on 1 January 2011.Other revenue from sale of merchan-dise, marketing rights, tickets, etc

Assumed to be received in the middle of each year from 2004 to 2015inclusive. The revenue from this source is expected to start at 0.1 perannum and increase each year by 0.1 up to and including 2011. The samerevenue is expected in 2012 as in 2011. After 2012 revenue is expected todecline by 0.2 per annum until 2015, after which year no further revenuewill be received from this source.

Revenue from sale of the stadiumand other infrastructure

This will be received on 1 January 2015

Determine the sale price of the stadium and other infrastructure that would have to be achieved for the project to beconsidered viable. (Institute/Faculty of Actuaries Examinations, September 2003) [3+3+11=17]

13. A company is considering investing in the following project. The company has to make an initial investment of3 payments, each of £105,000. The first is due at the start of the project, the second 6 months later, and the thirdpayment is due one year after the start of the project.After 15 years, it is assumed that a major refurbishment of the infrastructure will be required, costing £200,000. Theproject is expected to provide no income in the first year, an income received continuously of £20,000 in the secondyear, £23,000 in the third year, £26,000 in the fourth year and £29,000 in the fifth year. Thereafter, the income isexpected to increase by 3% per annum (compound) at the start of each year.The income is expected to cease at the end of the 30th year from the start of the project. The cash flow within eachyear is assumed to be received at a constant rate.

(i) Calculate the net present value of the project at a rate of interest of 8% p.a. effective.(ii) Show that there is no discounted payback period within the first 15 years, assuming an effective rate of interest

of 8% p.a.(iii) Calculate the discounted payback period for the project, assuming an effective rate of interest of 8% p.a.

(Institute/Faculty of Actuaries Examinations, May 1999. Specimen Examination.) [8+8+6=22]

There are further exercises on project appraisal in exercises 3 of chapter 3 on page 54.


7 Measuring investment performance7.1 Background. Measuring the performance of a fund brings special problems.

Example 7.1a. (a) An investor buys 1 share for £50 at time 0. At time 1, the share price is £53 and he also receivesa dividend of £2 from this share. He then buys a second share (for £53). At time 2, he receives a dividend of £2 foreach of these 2 shares and their price is then £54. Find the internal rate of return.(b) Now suppose that instead of buying just one extra share at time 1, the investor buys 10 shares. Hence his totalholding is then 11 shares. Find the internal rate of return.

Solution. The fund position can be summarised as follows.

Part (a): Time, t 0 1 2 Part (b): 0 1 2

Fund value at time t− 0 0 53 108 0 53 594Dividend receipts at time t 0 2 4 0 2 22

Investment at time t 50 53 0 50 530 0

Net cash flow (out) at time t 50 51 −4 50 528 −22

For part (a), the equation of value at t = 0 is

50 +51

1 + i− 4

(1 + i)2 =108

(1 + i)2

This quadratic equation in i has solution i = 0.0712 or 7.12%. For part (b), we have 6.02%.

For part (a) of the previous example, the investment of £50 at time 0 produced a gain of £5 in year 1. Thiscorresponds to 10%. The investment of £53 at time 1 produced a gain of £3 in year 2. This corresponds to5.66%. The internal rate of return is 7.12% and this is also called the money weighted rate of return (MWRR).It is influenced more by the performance in year 2 when 2 shares are held than the performance in year 1 whenonly 1 share is held. For part (b), the MWRR is even lower at 6.02% because 11 shares are held at time 2.

The MWRR is not a good indicator of the performance of an investment fund because it is sensitive to theamounts and timings of cash flows, which the fund manager does not control. Also, the equation for theMWRR may not have a unique or indeed any solution.

The time weighted rate of return (TWRR) ignores the number of shares and is calculated as follows:In year 1 the rate of return is i1 = 0.1 and in year 2 the rate of return is i2 = 0.0566. The TWRR, iT is definedby

(1 + iT )2 = (1 + i1)(1 + i2)In our particular case, this becomes

iT =

√5550

112106− 1 = 0.0781

The value of the TWRR is 7.81% for both parts (a) and (b).

7.2 General results. Consider the following general situation:Time, t 0 t1 t2 · · · tnFund value at time t− 0 f0−0 ft1−0 ft2−0 · · · ftn−0Net cash flow at time t c0 ct1 ct2 · · · ctn

Fund value at time t + 0 f0 = f0−0 + c0 ft1 = ft1−0 + ct1 ft2 = ft2−0 + ct2 · · · ftn = ftn−0 + ctn

If t > tn and ft is the value of the fund at time t, then the equation of value for the MWRR isf0(1 + i)t + ct1(1 + i)t−t1 + · · · + ctn(1 + i)t−tn = ft

The time weighted rate of return (TWRR) is given by

(1 + i)t =ft1−0

f0

ft2−0

ft1· · · ftn−0

ftn−1

ftftn

In general, the TWRR is more appropriate for assessing the performance of a fund manager who has no controlover the timings and sizes of investments. The MWRR is more appropriate for measuring the performance ofan individual investor who does determine the sizes and timings of investments.


7.3 The linked internal rate of return (LIRR). The TWRR measure also has disadvantages: it requiresknowledge of all the cash flows and the fund values at all the cash flow dates.

The linked internal rate of return (LIRR) is a combination of the MWRR and TWRR approaches: the MWRR(equivalently, the internal rate of return) is calculated at frequent intervals and then the TWRR formula is usedto combine these quantities.

In more detail: suppose t0 = 0 < t1 < · · · < tn and the annual effective rate of interest earned by the fund inthe interval (tr−1, tr) is ir. Then the LIRR is i where

(1 + i)tn = (1 + i1)t1(1 + i2)t2−t1(1 + i3)t3−t2 · · · (1 + in)tn−tn−1

The LIRR is useful if the fund is not valued every time there is a cash flow. The value of LIRR can then beused as an approximation to the unknown TWRR.The values of LIRR and TWRR will be the same if the lengths of the intervals (tr−1, tr) are the same in bothcalculations.

The method of calculation is best understood from an example.

Example 7.3a. An investment fund is valued once a year at the beginning of April. You are given the followinginformation about its performance:

Date Value of fund Deposits

April 1, 1996 £50,000October 1, 1996 £20,000April 1, 1997 £71,000April 1, 1998 £97,000

Find the linked internal rate of return for the period of April 1, 1996 to April 1, 1998 using subintervals of a year.(University of Warwick ST334 Examinations, 2012)

Solution. Use units of £1,000. Let i1 denote the annual effective rate of interest earned by the fund in the intervalfrom April 1, 1996 to April 1, 1997. Then 50(1 + i1) + 20(1 + i1)1/2 = 71. This is the quadratic 50x2 + 20x− 71 = 0which has solution x = 1.00830. Hence 1 + i1 = x2 = 1.01666889. Also 1 + i2 = 97/71.Hence if iL denotes the LIRR, we have (1 + iL)2 = 1.01666889× 97/71 and hence 1 + iL = 1.17854578768. Hencethe LIRR is 17.86%.

7.4

Summary.• Money Weighted Rate of Return (MWRR): same as internal rate of return.

f0(1 + i)t + ct1(1 + i)t−t1 + · · · + ctn(1 + i)t−tn = ft• Time Weighted Rate of Return (TWRR): requires value of fund at times of all cash flows.

(1 + i)t =ft1−0

f0

ft2−0

ft1· · · ftn−0

ftn−1

ftftn

(7.4a)

= (1 + i1)t1(1 + i2)t2−t1 · · · (1 + in)t−tn−1

• Linked Internal Rate of Return (LIRR): used if value of fund not known at times of all cash flows.


1. The following table gives information concerning an investment fund.

Calendar Year 1997 1998 1999 2000£ millions £ millions £ millions £ millions

Value of fund at 30 June — 460 500 650Net cash flow received on 1 July — 50 40 60Value of fund at 31 December 400 550 600 X

If the time weighted rate of return on the fund during the period from 31 December 1997 to 31 December 2000 is11% per annum effective, calculate X , the value of the fund on 31 December 2000.



2. The following table gives information concerning an investment fund.

Calendar Year 2000 2001 2002£ million £ million £ million

Value of fund on 1 January before cash flow 100 80 200Net cash flow received on 1 January 20 30 10Value of fund on 31 December 80 200 200

Calculate the effective time weighted rate of return over the three year period.(Institute/Faculty of Actuaries Examinations, September 2003) [3]

3. You are given the following information in respect of a pension fund:

Calendar Value of fund at Value of fund at Net cash flow received onYear 1 January 30 June 1 July

1997 £180,000 £212,000 £25,0001998 £261,000 £230,000 £18,0001999 £273,000 £295,000 £16,0002000 £309,000

Calculate the annual effective time weighted rate of return earned on the fund over the period from 1 January 1997to 1 January 2000. (Institute/Faculty of Actuaries Examinations, April 2000) [4]

4. A fund has a value of £120,000 on 1 January 2001. A net cash flow of £20,000 was received on 1 November 2001and a further net cash flow of £48,000 was received on 1 May 2002. Immediately before receipt of the first cash flow,the fund had a value of £137,000 and immediately before the second cash flow the fund had a value of £173,000.The value of the fund on 31 December 2002 was £205,000.

(i) Calculate the annual effective time weighted rate of return earned on the fund for the period 1 January 2001 to31 December 2002.

(ii) Discuss the relative strengths and weaknesses of using the time weighted rate of return as opposed to the moneyweighted rate of return when comparing the performance of two investment managers over the same period.


5. An investment fund had a market value of £2.2 million on 31 December 2001 and £4.2 million on 31 December 2004.It had received a net cash flow of £1.44 million on 31 December 2003.The money weighted rate of return and the time weighted rate of return for the period from 31 December 2001 to31 December 2004 are equal (to two decimal places).Calculate the market value of the fund immediately before the net cash flow on 31 December 2003.

(Institute/Faculty of Actuaries Examinations, April 2005 ) [7]

6. A fund had a value of £21,000 on 1 July 2003. A net cash flow of £5,000 was received on 1 July 2004 and a furthercash flow of £8,000 was received on 1 July 2005. Immediately before receipt of the first net cash flow, the fund hada value of £24,000, and immediately before receipt of the second net cash flow the fund had a value of £32,000. Thevalue of the fund on 1 July 2006 was £38,000.

(i) Calculate the annual effective money weighted rate of return earned on the fund over the period 1 July 2003 to1 July 2006.

(ii) Calculate the annual effective time weighted rate of return earned on the fund over the period 1 July 2003 to1 July 2006.

(iii) Explain why the values in (i) and (ii) differ.(Institute/Faculty of Actuaries Examinations, April 2007) [3+3+2=8]

7. A life insurance fund had assets totalling £600m on 1 January 2003. It received net income of £40m on 1 Jan-uary 2004 and £100m on 1 July 2004. The value of the fund was:

£450m on 31 December 2003; £500m on 30 June 2004; and £800m on 31 December 2004.(i) Calculate, for the period 1 January 2003 to 31 December 2004, to three decimal places:

(a) the time weighted rate of return.(b) the linked internal rate of return, using subintervals of a calendar year.

(ii) Explain why the linked internal rate of return is higher than the time weighted rate of return.(Institute/Faculty of Actuaries Examinations, September 2006) [8+2=10]

8. A pension fund had assets totalling £40 million on 1 January 2000. It received net income of £4 million on 1 Jan-uary 2001 and £2 million on 1 July 2001. The value of the fund totalled:

£43 million on 31 December 2000; £49 million on 30 June 2001 and £53 million on 31 December 2001.(i) Calculate for the period 1 January 2000 to 31 December 2001, to 3 decimal places:

(a) the time weighted rate of return per annum


(b) the linked internal rate of return, using sub-intervals of a calendar year.(ii) State both in general, and in this particular case, when the linked internal rate of return will be identical to the

time weighted rate of return. (Institute/Faculty of Actuaries Examinations, April 2003) [3+5+2=10]

9. A pension fund makes the following investments(£m):

1 January 2004 1 July 2004 1 January 2005 1 January 200612.5 6.6 7.0 8.0

The rates of return on money invested in the fund were as follows:

1 January 2004 to 1 July 2004 to 1 January 2005 to 1 January 2006 to30 June 2004 31 December 2004 31 December 2005 31 December 2006

5% 6% 6.5% 3%You may assume that 1 January to 30 June and 1 July to 31 December are precise half-year periods.

(i) Calculate the linked internal rate of return per annum over the 3 years from 1 January 2004 to 31 December 2006,using semi-annual sub-intervals.

(ii) Calculate the time weighted rate of return per annum over the 3 years from 1 January 2004 to 31 December 2006.(iii) Calculate the money weighted rate of return per annum over the 3 years from 1 January 2004 to 31 Decem-

ber 2006.(iv) Explain the relationship between your answers to (i), (ii) and (iii) above.


10. The value of the assets held by an investment fund on 1 January 2012 was £1.3 million.

On 30 September 2012, the value of the assets was £1.9 million.On 1 October 2012, there was a net cash outflow from the fund of £0.9 million.On 31 December 2012, the value of the assets was £0.8 million.

(i) Calculate the annual effective time-weighted rate of return (TWRR) for 2012.(ii) Calculate the annual effective money-weighted rate of return (MWRR) for 2012 to the nearest 1%.

(iii) Explain why the MWRR is significantly higher than the TWRR.(Institute/Faculty of Actuaries Examinations, April 2013) [2+3+2=7]

11. The value of the assets held by an investment fund on 1 January 2011 was £2.3 million. On 30 April 2011, the valueof the assets had risen to £2.9 million and, on 1 May 2011, there was a net cash inflow to the fund of £1.5 million.On 31 December 2011, the value of the assets was £4.2 million.

(i) Calculate the annual effective time-weighted rate of return (TWRR) for 2011.(ii) Calculate, to the nearest 0.1%, the annual effective money-weighted rate of return (MWRR) for 2011.

(iii) Explain why the TWRR is significantly higher than the MWRR for 2011.(Institute/Faculty of Actuaries Examinations, April 2012) [2+4+2=8]

12. In an accumulation fund, income is retained and used to increase the value of the units. Suppose the price in penceof the units in two accumulation funds, fund A and fund B, at various dates was as follows:

Date 1/4/2001 1/4/2002 1/4/2003 1/4/2004 1/4/2005 1/4/2006Fund A: unit price 80 90 98 180 190 160Fund B: unit price 70 67 66 100 110 130

Suppose the rate of inflation3 is given by the index in the following table:

Date 1/4/2001 1/4/2002 1/4/2003 1/4/2004 1/4/2005 1/4/2006Inflation index 100 102 105 117 120 122

An individual invests £1000 in fund A on 1/4/2001 and a further £1000 in fund A on 1/4/2002. On 1/4/2003 hewithdraws £1000 from fund A and invests this £1000 in fund B.(a) Calculate the real time weighted rate of return of the total investment for the period 1/4/2001 to 1/4/2006.(b) Derive an equation for the real money weighted rate of return of the total investment for the period 1/4/2001 to

1/4/2006. Use this equation to determine whether the real money weighted rate of return is greater or less thanthe real time weighted rate of return. (University of Warwick ST334 Examinations, 2014)

13. An investment fund is valued at £120 million on 1 January 2010 and at £140 million on 1 January 2011. Immediatelyafter the valuation on 1 January 2011, £200 million is paid into the fund. On 1 July 2012, the value of the fund is£600 million.

(i) Calculate the annual effective time-weighted rate of return over the two-and-a-half year period.(ii) Explain why the money weighted rate of return would be higher than the time weighted rate of return.


3 Real and money rates of interest are considered in section 4 of chapter 5 on page 91.


14. (i) State the strengths and weaknesses of using the money weighted rate of return as opposed to the time weightedrate of return as a measure of an investment manager’s skill.

(ii) An investor had savings of £41,000 in an account on 1 January 2006. He invested a further £12,000 in thisaccount on 1 August 2006. The total value of the account was £45,000 on 31 July 2006 and was £72,000 on31 December 2007.Assuming that the investor made no further deposits or withdrawals in relation to this account, calculate theannual effective time weighted rate of return for the period 1 January 2006 to 31 December 2007.


CHAPTER 3

Perpetuities, Annuities & Loans

1 Perpetuities

1.1 Perpetuities with constant payments. A perpetuity gives a fixed amount of income every year forever.Such securities are rare but do exist—for example, consols1 issued by the UK government are undated.

Suppose the rate of interest per unit time is constant and equal to i. Let ν = 1/(1 + i) denote the value at time0 of the amount 1 at time 1. The value at time 0 of the following cash-flow sequence:

Time 0 1 2 3 . . .Cash flow, c 0 1 1 1 . . .

↑ ↑a∞ a∞

is denoted a∞ or a∞ ,i and is given by

a∞ = ν + ν2 + ν3 + · · · =ν

1− ν=

1i

(1.1a)

The value of this perpetuity at time 1 is denoted a∞ or a∞ ,i and is given by

a∞ = (1 + i)a∞ = 1 + ν + ν2 + ν3 + · · · =1

1− ν=

1d

=1 + ii

(1.1b)

where d = 1− ν is the discount rate.

Example 1.1a. Assume a University post costs £50,000 per year and that the interest rate is 4% per annum. Howmuch must a benefactor give a University in order to establish a permanent post?

Solution. Using the above formula gives

a∞ =50,0000.04

= 1,250,000

1.2 In advance and in arrears. Suppose the rate of interest per unit time is constant and equal to i. Letν = 1/(1 + i) denote the value at time 0 of the amount 1 at time 1. Hence ν = 1 − d where d is the discountrate.

If A borrows £(1 − d) at time 0, then he has to repay £(1 − d)(1 + i) = 1 at time 1. This can be interpreted asfollows:

• The principal £(1−d) is returned at time 1 together with the interest £i(1−d) = £d. The interest is paidin arrears.OR• £1 is borrowed at time 0. The interest £d is paid in advance and the principal of £1 is returned at time 1.The interest £d paid at time 0 is the present value at time 0 of the quantity i at time 1. Hence d = i/(1 + i).

Thus a∞ can be regarded as the value of a perpetuity when payments are made in arrears and a∞ can beregarded as the value of a perpetuity when payments are made in advance.

1 These were redeemed in 2015.



1.3 Increasing perpetuities. Consider the following increasing perpetuity:

Time 0 1 2 3 . . .Cash flow, c 0 1 2 3 . . .

↑ ↑(Ia)∞ (Ia)∞

The value at time 0 is denoted (Ia)∞ or (Ia)∞ ,i and is given by

(Ia)∞ = ν + 2ν2 + 3ν3 + · · · =ν

(1− ν)2 =1id

(1.3a)

The value of the perpetuity at time 1 is denoted (Ia)∞ or (Ia)∞ ,i and is given by

(Ia)∞ = (1 + i)(Ia)∞ = 1 + 2ν + 3ν2 + 4ν3 + · · · =1

(1− ν)2 =1d2 (1.3b)

where, as usual, d = i/(1 + i). Clearly, (Ia)∞ = ν (Ia)∞ .

2 Annuities2.1 Annuities with constant payments: present values. Suppose n ≥ 1 and the payment 1 is received attime points 1, 2, . . . , n.

Time 0 1 2 · · · n− 1 nCash Flow, c 0 1 1 · · · 1 1

↑ ↑an an

(2.1a)

The value of the cash flow c in the display (2.1a) at time 0 is denoted an or an,i and is called the presentvalue of the immediate annuity-certain a.2 This corresponds to a sequence of payments made in arrears.

If i = 0 then clearly an = n. Otherwise,

an = ν + ν2 + · · · + νn =ν(1− νn)

1− ν=

1− νn

iThe quantity an is widely tabulated in actuarial tables, and so it is often useful to express other formulæ interms of this quantity. The definition is extended to all n ≥ 0 as follows:

Definition 2.1a. Suppose n ≥ 0 and i ≥ 0. Let ν = 1/(1 + i). Then

an,i =

n if i = 01− νn

iif i 6= 0

(2.1b)

The value of the cash flow c in the display (2.1a) at the time of the first payment is denoted an or an,i and iscalled the present value of the annuity-due a. This corresponds to a sequence of payments made in advance.

If i = 0 then clearly an = n. Otherwise,

an = (1 + i)an = 1 + ν + ν2 + · · · + νn−1 =1− νn

1− ν=

1− νn

d(2.1c)

This leads to the following definition:

Definition 2.1b. Suppose n ≥ 0 and i ≥ 0. Then an,i = (1 + i) an,i

Clearly if i > 0 then

a∞ = limn→∞

an =1i

and a∞ = limn→∞

an =1d

2 The word certain is used to distinguish it from annuities which cease to pay out when the holder dies.

3 Perpetuities, Annuities and Loans Sep 27, 2016(9:51) Section 2 Page 45

Example 2.1c. Consider a loan of £1,000 at 12% per annum compounded monthly. Suppose the loan is repaid over5 years by 60 equal monthly payments in arrears. How much are the monthly payments?

Solution. The interest rate is 0.01 per month. Hence we need the value of an,i with n = 60 and i = 0.01. Usingstandard tables gives a60 ,0.01 = 44.955038 and so the monthly payment is

1000a60 ,0.01

= 22.25

Or from first principles: let x denote the required payment, then

1000 = xν(1− ν60)

1− ν= x

1− ν60

iand so x =

1000i1− ν60 = 22.25

The process of gradually repaying a debt is called amortisation.

2.2 Annuities with constant payments: accumulated values.The value of the cash flow c in the display (2.1a) at time n, the time of the last payment, is denoted sn or sn,iand is called the accumulated value of the immediate annuity-certain a.

The value of the cash flow c in the display (2.1a) at time n+ 1, one time unit after the time of the last payment,is denoted sn or sn,i and is called the accumulated value of the immediate annuity-due a.

If i = 0 then sn = sn = n. Otherwise,sn = (1 + i)n−1 + (1 + i)n−2 + · · · + 1

= (1 + i)nan =(1 + i)n − 1

i(2.2a)

andsn = (1 + i)n + (1 + i)n−1 + · · · + (1 + i)

= (1 + i)nan = (1 + i)(1 + i)n − 1

i=

(1 + i)n − 1d

(2.2b)

Time 0 1 · · · n n + 1Cash Flow, c 0 1 · · · 1 0

↑ ↑sn sn

It follows that sn is the accumulated value at time n of the annuity if payments are made in arrears, whilstsn is the accumulated value at time n if the n payments are made in advance at times 0, 1, . . . , n − 1. Thedefinitions are extended to all n ≥ 0 as follows:

Definition 2.2a. Suppose n ≥ 0 and i ≥ 0. Thensn,i = (1 + i)n an,i and sn,i = (1 + i)n an,i

2.3 Annuities payable mthly. Instead of a payment of 1 every year, suppose payments of 1/m are paidm times per year. So the cash flow is as follows:

Time 0 1/m 2/m · · · (nm− 1)/m nm/m (nm + 1)/mCash Flow, c 0 1/m 1/m · · · 1/m 1/m 0

↑ ↑ ↑ ↑a(m)n a(m)

n s(m)n s(m)

n

(2.3a)

Present value of the annuity in display (2.3a). Denote the present value (time 0) of this annuity by a(m)n or

a(m)n,i . We suppose m and n are positive integers and let ν = 1/(1 + i). Then a(m)

n,i = n if i = 0, and

a(m)n = NPV(c) =

1m

nm∑j=1

νj/m =1m

ν1/m(1− νn)1− ν1/m

=1m

1− νn

(1 + i)1/m − 1=

1− νn

i(m) if i 6= 0.


We can derive this result another way. Recall that the following 4 sequences of payments are equivalent (seechapter 1).

Time 0 1/m 2/m · · · (m− 2)/m (m− 1)/m 1d

d(m)/m d(m)/m d(m)/m · · · d(m)/m d(m)/m

i(m)/m i(m)/m · · · i(m)/m i(m)/m i(m)/m

i

Hence the sequence ofm equal payments of i(m)/m at times 1/m, 2/m, . . . , 1 is equivalent to a single paymentof i at time 1. Hence the sequence of nm equal payments of 1/m at times 1/m, 2/m, . . . , n is equivalent to nequal payments of i/i(m) at times 1, 2, . . . , n. Using the result an = (1 − νn)/i for a sequence of paymentsof 1 at times 1, 2, . . . , n gives

a(m)n,i =

i

i(m) an,i (2.3b)

Equation (2.3b) is a very important result—it enables fast calculation of a(m)n,i for standard values of i, m and n

by using tabulated values of i/i(m) and an,i.

Example 2.3a. Find the present value of an annuity for 9 years payable quarterly if the effective interest rate is 7%per annum.

Solution. Use the formula: a(4)9 =

i

i(4) a9 . The quantityi

i(4) is given in the tables for 7%—it is in the constants on theleft hand side of the page and is given as 1.025 880. The quantity a9 is given in the main table as 6.515 232. Hencethe present value of our annuity is 1.025 880× 6.515 232.

Similarly, a(m)n or a(m)

n,i are used to denote the present value at the time of the first payment. We can work outthe formula for this quantity from first principles as follows:

a(m)n =

1m

nm−1∑j=0

νj/m =1m

1− νn

1− ν1/m =1− νn

d(m) (2.3c)

An alternative method is to use the fact that the cash flow is equivalent to the sequence of n equal payments ofd/d(m) at the times 0, 1, . . . , n− 1. Then use an = (1− νn)/d for a sequence of equal payments of size 1 attimes 0, 1, . . . , n− 1. Hence

a(m)n =

d

d(m) an =1− νn

d(m) =i

d(m) an

The above definitions are generalised as follows:

Definition 2.3b. Suppose n ≥ 0, m ≥ 0 and i ≥ 0. Let ν = 1/(1 + i). Then

a(m)n,i =

n if i = 0

1− νn

i(m) if i 6= 0and a(m)

n,i = (1 + i)1/m a(m)n,i

For example,• If m = 4 and n = 21/4, then a(m)

n denotes 21 payments each with value 1/4 at times 1/4, 2/4, . . . , 21/4.• If m = 1/4 and n = 32 then a(m)

n denotes 8 payments each with value 4 at times 4, 8, . . . , 32.• If nm is not an integer, then the formula has no sensible interpretation. For example, if n = 2.75 and m = 2,then we could consider an annuity of 5 six-monthly payments of 1/2 each followed by a payment of 1/4 after2.75 years; i.e. a(2)

2.5 + 14 ν

2.75. But this is not the value of the formula in definition 2.3b.

Accumulated value in display (2.3a). Let s(m)n denote the value of the annuity at the time of the last payment

and let s(m)n denote the value of the annuity at time n + 1/m, which is one payment interval after the time of

the last payment.


Definition 2.3c. Suppose n ≥ 0, m ≥ 0 and i ≥ 0. Thens(m)n,i = (1 + i)n a(m)

n,i and s(m)n,i = (1 + i)n a(m)

n,i (2.3d)

Note that a(1)n,i = an,i, a(1)

n,i = an,i, s(1)n,i = sn,i and s(1)

n,i = sn,i

Generalisation. Suppose we have the following cash flow:

Time 0 1 2 · · · n− 1 nCash Flow, c 0 x1 x2 · · · xn−1 xn

The net present value (time 0) is given by:

NPV(c) =n∑j=1

νjxj

Now suppose that the payment xj is replaced by m equal payments of xj/m at times j − 1 + 1/m, j − 1 +2/m, . . . , j for every j = 1, 2, . . . , n. Then the new net present value is

i

i(m)

n∑j=1

νjxj

2.4 Deferred annuities. Consider the following cash flow (where k and n are positive integers):

Time 0 1 · · · k k + 1 · · · k + n− 1 k + nCash Flow, c 0 0 · · · 0 1 · · · 1 1

This can be considered as an immediate annuity-certain delayed for k time units. Its present value (time 0) isdenoted

NPV(c) = k|an =k+n∑j=k+1

νj

Clearly we have

k|an = ak+n − ak

= νkan

The value at time 1 of the above annuity is termed the value of the annuity-due delayed for k time units. Thisvalue is denoted k|an and is

NPV(c; t = 1) = k|an =k+n−1∑j=k

νj

Clearly we have

k|an = νkan

Deferred annuities payable mthly. Consider the following cash flow: k|an and is

Time 0 · · · k k + 1/m k + 2/m · · · k + (nm− 1)/m k + nm/mCash Flow 0 · · · 0 1/m 1/m · · · 1/m 1/m

The value at time 0 of this cash flow is denoted k|a(m)n and the value at time 1/m is denoted k|a

(m)n . This leads

to the following definitions:

Definition 2.4a. Suppose k ≥ 0, n ≥ 0, m ≥ 0 and i ≥ 0 and ν = 1/(i + 1). Then

k|a(m)n = νka(m)

n and k|a(m)n = νka(m)

n

The following are immediate consequences of the definition:

k|a(m)n = νk−1/ma(m)

n and k|a(m)n = a(m)

n+k − a(m)k

and k|a(m)n = a(m)

n+k − a(m)k

The derivations are left as exercises.


2.5 Increasing annuities. Consider the following increasing annuity:

Time 0 1 2 · · · n n + 1Cash Flow, c 0 1 2 · · · n 0

↑ ↑ ↑ ↑(Ia)n (Ia)n (Is)n (Is)n

The present value (time 0) of this annuity is denoted (Ia)n and the present value at time 1 is denoted (Ia)n .Using the obvious notation, the accumulated values are (Is)n = (1 + i)n(Ia)n and (Is)n = (1 + i)n(Ia)n .Clearly:

(Ia)n = NPV(c) = ν + 2ν2 + · · · + nνn

=1i

[1− νn

1− ν− nνn

](2.5a)

and so

(Ia)n =an − nνn

i=

an − nνn+1

1− ν(2.5b)

and(Ia)n = NPV(c; t = 1) = 1 + 2ν + 3ν2 + · · · + nνn−1

= (1 + i)(Ia)n

=an − nνn

1− ν=

an − nνn+1

ν(1− ν)(2.5c)

We can derive these results by considering equivalent cash flows:

Time 0 1 2 · · · n n + 1 n + 2 · · ·Cash Flow, c 0 1 2 · · · n 0 0 · · ·Cash Flow, c1 0 1 2 · · · n n + 1 n + 2 · · ·Cash Flow, c2 0 0 0 · · · 0 n n · · ·Cash Flow, c3 0 0 0 · · · 0 1 2 · · ·

Clearly

NPV(c; t = 0) = NPV(c1; t = 0)− NPV(c2; t = 0)− NPV(c3; t = 0)= (Ia)∞ − nνna∞ − νn(Ia)∞

and hence(Ia)n = (1− νn)(Ia)∞ − nνna∞

which gives equation (2.5a) by using the results (Ia)∞ = 1/i(1−ν) and a∞ = 1/i on pages 43 and 44 above.Similarly

(Ia)n = NPV(c; t = 1)= NPV(c1; t = 1)− NPV(c2; t = 1)− NPV(c3; t = 1)= (1− νn)(Ia)∞ − nνna∞

Here is another decomposition for an increasing annuity:

Time 0 1 2 3 · · · nCash Flow, c 0 P P +Q P + 2Q · · · P + (n− 1)QCash Flow, c1 0 P −Q P −Q P −Q · · · P −QCash Flow, c2 0 Q 2Q 3Q · · · nQ

HenceNPV(c) = (P −Q)an +Q(Ia)n


Consider the following cash flow:

Time 0 1 2 3 · · · nCash Flow, c 0 2 3 4 · · · n + 1

Applying the previous result with P = 2 and Q = 1 shows that:NPV(c) = an + (Ia)n

and hence we obtain the following result:(Ia)n+1 = 1 + an + (Ia)n

2.6 Decreasing annuities. The term (Da)n denotes the present value of a decreasing annuity payable inarrears for n units of time with payments n, n− 1, . . . , 1. See exercises 28 and 27.

2.7 Continuously payable annuities with constant payments. Suppose the force of interest is constant andequal to δ. Hence the present value function is ν(t) = e−δt = νt and i = eδ − 1.

Present Value. Suppose n ≥ 0. and let an,i or an denote the value at time 0 of an annuity payable continu-ously between times 0 and n when the rate of payment ρ(t) is constant and equal to 1. Then an,i = n if i = 0,or equivalently, if δ = 0. Otherwise, by using 1 + i = eδ and ν = 1/(1 + i), we get

an,i =∫ n

0ρ(t)ν(t) dt =

∫ n

0νt dt =

νt

ln ν

∣∣∣∣nt=0

=1− νn

δ=i

δan,i if δ 6= 0 (2.7a)

The leads to the following definition:

Definition 2.7a. Suppose n ≥ 0 and δ ≥ 0. Let i = eδ − 1. Then

an,i =

n = an,i if i = 0

1− νn

δ=i

δan,i if i 6= 0

For a continuously payable perpetuity we have

a∞ =∫ ∞

0ρ(t)ν(t) dt =

∫ ∞0

e−δt dt =1δ

=i

δa∞ if i 6= 0 (2.7b)

Accumulated value. Let sn,i or sn denote the accumulated value of the annuity at time n, the time whenpayments cease. Then

sn =∫ n

0ρ(t)eδ(n−t) dt = eδn an = (1 + i)n an (2.7c)

leading to the definition:

Definition 2.7b. Suppose n ≥ 0 and i ≥ 0. Then sn,i = (1 + i)n an,i

Clearly

sn,i =

an,i = n if i = 0

(1 + i)n − 1δ

=i

δsn if i 6= 0

Deferred continuously payable annuities. Let k|an denote the present value at time 0 of an annuity of 1 pertime unit over the interval (k, k + n). Then

k|an =∫ k+n

kρ(t)ν(t) dt =

∫ k+n

ke−δt dt

=∫ k+n

0e−δt dt−

∫ k

0e−δt dt

= ak+n − ak

= νkan


2.8 Increasing continuously payable annuities. There are two common payment rates, ρ:

• the step function ρ(t) = j for t ∈ (j − 1, j] and j = 1, 2, . . . , n;• the continuous function ρ(t) = t for t ∈ (0, n).

For the step function, it can be shown (exercise 25) that:

(Ia)n =∫ n

0ρ(t)ν(t) dt =

n∑j=1

∫ j

j−1jνt dt

=an − nνn

δwhere δ = − ln ν. (2.8a)

The notation for the accumulated value is (Is)n = (1 + i)n(Ia)n .

For the continuous function ρ(t) = t, we have (see exercise 35)

(Ia)n =∫ n

0ρ(t)ν(t) dt =

∫ n

0tνt dt =

an − nνn

δ(2.8b)

For this case, the notation for the accumulated value is (Is)n = (1 + i)n(Ia)n .

2.9

Summary.• Annuities with constant payments payable once per year in arrears for n years (an annuity-certain).

an = ν + ν2 + · · · + νnNet present value at time 0:

=1− νn

iif i 6= 0.

sn = (1 + i)nanAccumulated value at time n:

• Annuities with constant payments payable once per year for n years in advance (an annuity-due).an = (1 + i)anNet present value at time 0:

sn = (1 + i)nanAccumulated value at time n + 1:

• Annuities with constant payments payable m times per year for n years with nm payments at times1/m, 2/m, . . . , nm/m. (Set m = 1 to get results for an , an , sn , and sn .)

a(m)n =

i

i(m) an if i 6= 0Net present value at time 0:

a(m)n = (1 + i)1/ma(m)

nNet present value at time 1/m:

s(m)n = (1 + i)na(m)

nAccumulated value at time n:

s(m)n = (1 + i)na(m)

nAccumulated value at time n + 1/m:

• Perpetuity with constant payments of 1 at times 1, 2, . . . .

a∞ = limn→∞

an =1i

Net present value at time 0:

a∞ = (1 + i)a∞Net present value at time 1:

• Increasing Annuities.

(Ia)n =an − nνn+1

1− νand (Ia)n = (1 + i)(Ia)n

• Increasing Perpetuities.

(Ia)∞ =1

i(1− ν)and (Ia)∞ = (1 + i)(Ia)∞

• Deferred annuities, k|an , and decreasing annuities, (Da)n .. . . continued

3 Perpetuities, Annuities and Loans Sep 27, 2016(9:51) Exercises 3 Page 51

Summary (continued).• Continuously payable annuity with constant rate δ over the interval (0, n).

anNet present value at time 0:sn = (1 + i)n anAccumulated value at time n:

• Increasing continuously payable annuities:(Ia)n for the step function ρ(t) = j for t ∈ (j − 1, j] and j = 1, 2, . . . , n,(Ia)n for ρ(t) = t.


1. Prove the following results:(a) sn = (1 + i)sn (b) sn+1 = sn + 1 (c) an = an−1 + 1 for n ≥ 2(d) ian + νn = 1 (e) dan + νn = 1 (f) isn + 1 = (1 + i)n

(g) dsn + 1 = (1 + i)n

Give interpretations of these results: for example, (a) is the result that sn is the value of the same series of paymentsas sn but 1 time unit later.

2. The force of interest, δ, is 6%. Find the value of (Ia)10 .(Institute/Faculty of Actuaries Examinations, September 1999) [3]

3. What is the present value of an annuity of £160 per annum payable quarterly in arrears for 4 years, at a rate of interestof 12% per annum, convertible half-yearly. (Institute/Faculty of Actuaries Examinations, April 1998) [3]

4. The annual effective rate of interest is 7%. Find the value of s(12)12 .


5. What is the value of s(4)20 at an effective rate of interest is 71/2% per annum.


6. An annuity due starts at £200 per year and thereafter decreases by £10 per year. After the payment of £130 hasbeen made, further payments cease. Express the present value, x, of the annuity in terms of a8 and (Ia)8 . Hencedetermine x if the effective interest rate is 7% per annum.

(Institute/Faculty of Actuaries Examinations, April 1999, adapted)

7. An investment project requires an initial investment of £100m. It is expected to yield an income of £20m per annumannually in arrears. At a rate of interest of 15% per annum effective, find the discounted payback period.


8. A series of payments is to be received annually in advance. The first payment will be £10. Thereafter, paymentsincrease by £2 per annum. The last payment will be made at the beginning of the tenth year. Which of the followingare equal to the present value of the annuity?

(I)∑9t=0 8νt + 2

∑9t=0 tν

t (II) 10a10 + 2(Ia)9 (III) 8a10 + 2(Ia)10(Institute/Faculty of Actuaries Examinations, September 1998) [3]

9. An annuity certain with payments of £150 at the end of each quarter is to be replaced by an annuity with the sameterm and present value, but with payments at the beginning of each month instead.Calculate the revised payments, assuming an annual force of interest of 10%.


10. A variable annuity, payable annually in arrears for n years is such that the payment at the end of year t is t2. Showthat the present value of the annuity can be expressed as

2(Ia)n − an − n2νn+1

1− ν(Institute/Faculty of Actuaries Examinations, April 2000. Part question.) [4]

11. An investor is to make payments of £100 annually in arrears for 6 years followed by £200 per annum payablehalf yearly in arrears for a further 6 years. The investment earns interest at 8% per annum convertible half yearly.Calculate the accumulated amount at the end of 12 years.


12. An investor pays £400 every half-year in advance into a 25-year savings plan.Calculate the accumulated fund at the end of the term if the interest rate is 6% per annum convertible monthly forthe first 15 years and 6% per annum convertible half-yearly for the final 10 years.


13. (i) Calculate s(12)5.5 at an effective rate of interest of 13% per annum.

(ii) Explain what your answer to (i) represents. (Institute/Faculty of Actuaries Examinations, April 2000) [5]

14. Recall that an = a(1)n , an = a(1)

n , sn = s(1)n and sn = s(1)

n .If i 6= 0, show that

a(m)n =

i

i(m) an a(m)n =

d

d(m) an s(m)n =

i

i(m) sn s(m)n =

d

d(m) sn =i

d(m) sn

15. (a) Suppose m and n are positive integers. Show thatam+n = am + νman

(b) A loan of £10,000 is to be repaid by constant monthly payments in arrear for 10 years. The interest rate is 1%per month effective. Find the monthly payment.

16. Suppose the force of interest is given by δ(t) = 0.02t for 0 ≤ t ≤ 5. Find s5 .

17. Suppose we have a perpetuity where the amount grows by a constant proportion every year. So we have the cashflow:

Time 0 1 2 3 . . .Cash flow, c 0 1 1 + g (1 + g)2 . . .

(a) Show that NPV(c) = 1/(i− g) for g < i.(b) Suppose the interest rate is 10%, and we want a perpetuity with initial payment 10,000 and which grows by 4%

per annum. Find the NPV.

18. Prove the following result:limm→∞

a(m)n = lim

m→∞a(m)n = an

19. Prove the following results:(a) i(m)s(m)

n = isn = dsn = d(m)s(m)n

(b) i(m)a(m)n = ian = dan = d(m)a(m)

n = δan . (All equal 1− νn.)

20. Show that

a(m)n,i =

(i

i(m) +i

m

)an,i

21. Suppose i ≥ 0, n ≥ 0, m > 0 and nm is an integer. Prove that

a(m)n,i =

1m

anm,j where j =i(m)

m

22. Perpetuity payable mthly. Let a(m)∞ denote the present value (at time 0) of a perpetuity where constant payments of

1/m are made m times per year in advance.

Time 0 1/m 2/m . . .

Cash flow 1/m 1/m 1/m . . .

Let a(m)∞ denote the present value (at time 0) of a perpetuity where constant payments of 1/m are made m times per

year in arrears.

Time 0 1/m 2/m . . .

Cash flow 0 1/m 1/m . . .

(a) Show that

a(m)∞ =

1d(m) and a(m)

∞ =1i(m)

(b) Hence show that1d(m) =

1m

+1i(m)


23. (a) Consider the following cash flow:

Time 0 r 2r 3r . . . kr

Cash flow, c 0 1 1 1 . . . 1

Show that the present value (time 0) of the cash flow c is

NPV(c) =akrsr

(b) Suppose n ≥ 0 and m = 1/r where r ≥ 1 is an integer. Suppose further that nm is an integer. Prove that

a(m)n =

1ms1/m

an

24. Suppose the interest rate is i, the number of payments per year is m and the number of increases per year is q whereq divides m. (For example, if m = 12 and q = 4, then payments are made monthly with quarterly increases.)

Time 0 1/m · · · 1/q − 1/m 1/q · · · 2/q − 1/m · · ·Cash Flow 1/(mq) 1/(mq) · · · 1/(mq) 2/(mq) · · · 2/(mq) · · ·

Denote the present value (time 0) by (I (q)a)(m)∞ . Show that

(I (q)a)(m)∞ = a(m)

∞ a(q)∞ =

1d(m)

1d(q)

25. Prove the following results about increasing continuously payable annuities:

(Ia)n =an − nνn

δand (Ia)n =

an − nνn

δ

26. Deferred annuities. Prove the following results:

(a) k|a(m)n = νk−1/ma(m)

n (b) k|a(m)n = a(m)

n+k − a(m)k

(c) k|a(m)n = a(m)

n+k − a(m)k

27. Decreasing Annuities. Consider the following cash flow:

Time 0 1 2 · · · nCash flow, c 0 n n− 1 · · · 1

Let (Da)n = NPV(c). Prove that (Da)n = (n + 1)an − (Ia)n .

28. A 20 year annuity certain provides payments annually in arrear of £200 in year 1, £180 in year 2, and so on, until thepayments have reduced to £60. Payments then continue at £60 per annum until the 20th payment under the annuityhas been made.Shown below are 3 suggested expressions for the present value of the annuity:

(I) 20(Da)10 − 20ν8(Da)3 + 60(a21 − a8 )(II) 200a20 − 20ν(Ia)7 − 140ν8a12

(III) 20(Da)7 + 60a20

Which of the expressions are correct? (Institute/Faculty of Actuaries Examinations, April 1998) [3]

29. An investor is considering the purchase of an interest in a warehouse for £250,000. The interest entitles the purchaserto receive rent for a 5 year period at a rate of £64,000 per annum quarterly in advance. Maintenance and managementcosts are also fixed for the 5 year period at a rate of £7,000 per annum payable by the investor at the beginning ofeach year. At the end of the 5 year period the purchaser neither makes nor receives any further payments.

(i) Calculate the annual effective rate of return.(ii) Calculate the annual effective real rate of return if inflation is 2% per annum for the 5 years.


30. An insurance company offers a customer two payment options in respect of an invoice for £456. The first optioninvolves 24 payments of £20 paid at the beginning of each month starting immediately. The second option involves24 payments of £20.50 paid at the end of each month starting immediately. The customer is willing to accept amonthly payment schedule if the annual effective interest rate per annum he pays is less than 5%.

Determine which, if any, of the payment options the customer will accept.(Institute/Faculty of Actuaries Examinations, September 2007) [4]


31. A financial regulator has brought in a new set of regulations and wishes to assess the cost of them. It intends toconduct an analysis of the costs and benefits of the new regulations in their first twenty years.

The costs are estimated as follows.• The cost to companies who will need to devise new policy terms and computer systems is expected to be

incurred at a rate of £50m in the first year increasing by 3% per annum over the twenty year period.• The cost to financial advisers who will have to set up new computer systems and spend more time filling in

paperwork is expected to be incurred at a rate of £60m in the first year, £19m in the second year, £18m in thethird year, reducing by £1m every year until the last year, when the cost incurred will be at the rate of £1m.

• The cost to consumers who will have to spend more time filling in paperwork and talking to their financialadvisers is expected to be incurred at a rate of £10m in the first year, increasing by 3% per annum over thetwenty year period.

The benefits are estimated as follows.• The benefits to consumers who are less likely to buy inappropriate policies is estimated to be received at a rate

of £30m in the first year, £33m in the second year, £36m in the third year, rising by £3m per year until the endof twenty years.

• The benefits to companies who will spend less time dealing with complaints from customers is estimated to bereceived at a rate of £12m per annum for twenty years.

Calculate the net present value of the benefit or cost of the regulations in their first twenty years at a rate of interestof 4% per annum effective. Assume that all costs and benefits occur continuously throughout the year.


32. A bank offers two repayment alternatives for a loan that is to be repaid over 10 years. The first requires the borrowerto pay £1,200 per annum quarterly in advance and the second requires the borrower to make payments at an annualrate of £1,260 every second year in arrears.Determine which terms would provide the best deal for the borrower at a rate of interest of 4% per annum effective.



δ(t) =

0.08 for 0 ≤ t ≤ 4;0.12− 0.01t for 4 < t ≤ 9;0.05 for t > 9.

(i) Determine the discount factor, ν(t), that applies at time t for all t ≥ 0.(ii) Calculate the present value at t = 0 of a payment stream, paid continuously from t = 10 to t = 12, under which

the rate of payment at time t is 100e0.03t.(iii) Calculate the present value of an annuity of £1,000 paid at the end of each year for the first 3 years.


34. A special type of loan is to be issued by a company. The loan is made up of 100,000 bonds, each of nominal valuee100. Coupons will be paid semi-annually in arrear at a rate of 4% per annum. The bonds are to be issued on1 October 2015 at a price of e100 per e100 nominal. Income tax will be paid by the bond holders at a rate of 25%on all coupon payments.Exactly half the bonds will be redeemed after ten years at e100 per e100 nominal. The bonds that are redeemedwill be determined by lot ( i.e. the bonds will be numbered and half the numbered bonds will be chosen randomlyfor redemption). Coupon payments on the remaining bonds will be increased to 7% per annum paid semi-annuallyand these bonds will be redeemed 20 years after issue at e130 per e100 nominal.An individual buys a single bond. Calculate, as an effective rate of return per annum:

(i) the maximum rate of return the individual can obtain from the bond;(ii) the minimum rate of return the individual can obtain from the bond;

(iii) the expected rate of return the individual will obtain from the bond.(iv) An investor is considering buying the whole loan. Show that the rate of return that this investor will obtain is

greater than the expected rate of return that the above individual who buys a single bond will receive.(Institute/Faculty of Actuaries Examinations, September 2015, corrected) [5+2+2+5=14]

Questions on project analysis.

35. (i) In an annuity, the rate of payment per unit of time is continuously increasing so that at time t it is t. Show thatthe value at time 0 of such an annuity payable until time n, which is represented by (Ia)n is given by

(Ia)n =an − nνn

δwhere δ is the force of interest per unit of time and an =

∫ n0 exp(−δt) dt.

(ii) A consortium is considering the purchase of a coal mine which has recently ceased production. The consortiumforecasts that:


(1) the cost of reopening the mine will be £700,000, and this will be incurred continuously throughout the firsttwelve months(2) after the first twelve months the revenue from sales of coal, less costs of sale and extraction, will growcontinuously from zero to £2,500,000 at a constant rate of £250,000 per annum(3) when the revenue from sales of coal, less costs of sale and extraction, reaches £2,500,000 it will then declinecontinuously at a constant rate of £125,000 per annum until it reaches £250,000(4) when the revenue declines to £250,000 production will stop and the mine will have zero value

Additional costs are expected to be constant throughout at £200,000 p.a. excluding the first year. These are alsoincurred continuously.

What price should the consortium pay to earn an internal rate of return (IRR) of 20% p.a. effective?(Institute/Faculty of Actuaries Examinations, September 1998) [14]

36. A property developer is constructing a block of offices. It is anticipated that the offices will take six months to build.The developer incurs costs of £40 million at the beginning of the project followed by £3 million at the end of eachmonth for the following six months during the building period. It is expected that rental income from the officeswill be £1 million per month, which will be received at the start of each month beginning with the seventh month.Maintenance and management costs paid be the developer are expected to be £2 million per annum payable monthlyin arrear with the first payment at the end of the seventh month. The block of offices is expected to be sold 25 yearsafter the start of the project for £60 million.

(i) Calculate the discounted payback period using an effective rate of interest of 10% per annum.(ii) Without doing any further calculations, explain whether your answer to (i) would change if the effective rate of

interest were less than 10% per annum.(Institute/Faculty of Actuaries Examinations, April 2007) [7+3=10]

37. An investor borrows £120,000 at an effective interest rate of 7% per annum. The investor uses the money to purchasean annuity of £14,000 per annum payable half-yearly in arrears for 25 years. Once the loan is paid off, the investorcan earn interest at an effective rate of 5% per annum on money invested from the annuity payments.

(i) Determine the discounted payback for this investment.(ii) Determine the profit the investor will have made at the end of the term of the annuity.


38. A piece of land is available for sale for £5,000,000. A property developer, who can lend and borrow money at arate of 15% per annum, believes that she can build housing on the land and sell it for a profit. The total cost ofdevelopment would be £7,000,000 which would be incurred continuously over the first two years after purchase ofthe land. The development would then be complete.The developer has three possible strategies. She believes that she can sell off the completed housing:• in three years time for £16,500,000;• in four years’ time for £18,000,000;• in five years’ time for £20,500,000.

The developer also believes that she can obtain a rental income from the housing between the time that the develop-ment is completed and the time of the sale,. The rental income is payable quarterly in advance and is expected to be£500,000 in the first year of payment. Thereafter, the rental income is expected to increase by £50,000 per annum atthe beginning of each year that the income is paid.

(i) Determine the optimum strategy if this is based upon using net present value as the decision criterion.(ii) Determine which strategy would be optimal if the discounted payback period were to be used as the decision

criterion.(iii) If the housing is sold in six years’ time, the developer believes that she can obtain an internal rate of return on

the project of 17.5% per annum. Calculate the sale price that the developer believes that she can receive.(iv) Suggest reasons why the developer may not achieve an internal rate of return of 17.5% per annum even if she

sells the housing for the sale price calculated in (iii).(Institute/Faculty of Actuaries Examinations, April 2006) [9+2+6+2=19]

39. (i) Explain what is meant by the following terms:(a) equation of value;(b) discounted payback period from an investment project.

(ii) An insurance company is considering setting up a branch in a country in which it has previously not operated.The company is aware that access to capital may become difficult in twelve years’ time. It therefore has twodecision criteria. The cash flows from the project must provide an internal rate of return greater than 9% perannum effective and the discounted payback period at a rate of interest of 7% per annum effective must be lessthan twelve years.The following cash flows are generated in the development and operation of the branch.Cash Outflows. Between the present time and the opening of the branch in three years’ time, the insurancecompany will spend £1.5m per annum on research, development and the marketing of products. This outlay isassumed to be a constant continuous payment stream. The rent on the branch building will be £0.3m per annum


paid quarterly in advance for twelve years starting in three years’ time. Staff costs are assumed to be £1m in thefirst year, £1.05m in the second year, rising by 5% per annum each year thereafter. Staff costs are assumed to beincurred at the beginning of each year starting in three years’ time and assumed to be incurred for twelve years.Cash Inflows. The company expects the sale of products to produce a net income at a rate of £1m per annumfor the first three years after the branch opens rising to £1.9m per annum in the next three years and to £2.5mfor the following six years. This net income is assumed to be received continuously throughout each year. Thecompany expects to be able to sell the branch operation 15 years from the present time for £8m.Determine which, if any, of the decision criteria the project fulfils.


40. An investor is considering investing in a capital project. The project requires a capital outlay of £500,000 at outsetand further payments at the end of each of the first 5 years, the first payment being £100,000 and each successivepayment increasing by £10,000. The project is expected to provide a continuous income at a rate of £80,000 in thefirst year, £83,200 in the second year and so on, with income increasing each year by 4% per annum compound. Theincome is received for 25 years.It is assumed that, at the end of 15 years, a further investment of £300,000 will be required and that the project canbe sold to another investor for £700,000 at the end of 25 years.

(i) Calculate the net present value of the project at a rate of interest of 11% per annum effective.(ii) Without doing any further calculations, explain how the net present value would alter if the interest rate had

been greater than 11% per annum effective.(Institute/Faculty of Actuaries Examinations, April 2008) [9+3=12]

41. A pension fund is considering investing in a major infrastructure project. The fund has been asked to make aninvestment of £2m for a 1% share in revenues from building a road. No other costs will be incurred by the pensionfund. The following revenues are expected to arise from the project:• In the first year, 40,000 vehicles a day will use the road, each paying a toll of £1.• In the second year, 50,000 vehicles a day will use the road, each paying a toll of £1.10.• In the third year, both the number of vehicles using the road and the level of tolls will rise by 1% from their level

in the second year. They will both continue to rise by 1% per annum compound until the end of the 20th year.• At then end of the 20th year, it is assumed that the road has no value as it will have to be completely rebuilt.You should assume that all revenue is received continuously throughout the year and that there are 365 days in allyears. Calculate the net present value of the investment in the road at a rate of interest of 8% per annum effective.


42. An investor is considering two projects, Project A and Project B. Project A involves the investment of £1,309,500in a retail outlet. Rent is received quarterly in arrear for 25 years, at an initial rate of £100,000 per annum. It isassumed that the rent will increase at a rate of 5% per annum compound, but with increases taking place every fiveyears. Maintenance and other expenses are incurred quarterly in arrear, at a rate of £12,000 per annum. The retailoutlet reverts to its original owner after 25 years for no payment.Project B involves the purchase of an office building for £1,000,000. The rent is to be received quarterly in advanceat an initial rate of £85,000 per annum. It is assumed that the rent will increase to £90,000 per annum after 20 years.There are no maintenance or other expenses. After 25 years the property reverts to its original owners for no payment.

(i) Show that the internal rate of return for Project A is 9% per annum effective.(ii) Calculate the annual effective internal rate of return for Project B. Show your working.

(iii) Discuss the extent to which the answers to parts (i) and (ii) above will influence the investor’s decision overwhich project to choose. (Institute/Faculty of Actuaries Examinations, April 2012) [5+4+3=12]

43. A property development company has just purchased a retail outlet for $4,000,000. A further $900,000 will be spentrefurbishing the outlet in 6 months’ time.An agreement has been made with a prospective tenant who will occupy the outlet beginning one year after thepurchase date. The tenant will pay rent to the owner for 5 years and will then immediately purchase the outlet fromthe property development company for $6,800,000. The initial rent will be $360,000 per annum. and this will beincreased by the same percentage compound rate at the beginning of each successive year. The rental income isreceived quarterly in advance.Calculate the compound percentage increase in the annual rent required to earn the company an internal rate of returnof 12% per annum effective. (Institute/Faculty of Actuaries Examinations, April 2015) [9]

44. A student has inherited £1m and is considering investing the money in two projects, A and B.Project A requires the investment of the whole sum in properties that are to be let out to tenants. The details are:• The student expects to receive an income from rents at an annual rate of £60,000 a year for 4 years after an initial

period of one year in which no income will be received.• Rents are expected to rise thereafter at the start of each year at a rate of 0.5% per annum.• The income will be received monthly in advance.


• The project involves costs of £10,000 per annum in the first year, rising at a constant rate of 0.5% per annum.• The costs will be incurred at the beginning of each year.• At the end of 20 years, the student expects to be able to sell the properties for £2m after which there will be no

further revenue or costs.

Project B involves the investment of the whole sum in an investment fund.• The fund is expected to pay an income of £60,000 per annum annually in advance and return the whole invested

sum at the end of 20 years.

(i) (a) Calculate the payback period for project B.(b) Show, by general reasoning or otherwise, that the payback period from project A is longer than that from

project B.

(ii) (a) Define the discounted payback period.(b) Determine the discounted payback period from project B at a rate of interest of 1% per annum effective.(c) Show, by general reasoning or otherwise, that the discounted payback period from project A is longer than

that from project B.

(iii) Determine the internal rate of return from project B expressed as an annual effective return.

(iv) Show that the internal rate of return from project A is higher than that from project B.

(v) Discuss which project is the better project given your answers to parts (i)(iv) above.(Institute/Faculty of Actuaries Examinations, September 2015) [5+5+3+10+3=26]

4 Loan schedules

4.1 Introduction.

Example 4.1a. Suppose A borrows £1,000 for 3 years at an effective interest rate of 7% per annum. Supposefurther that A repays the loan by equal amounts of x at the ends of years 1, 2 and 3. Find x.

Solution. The cash flow is as follows:

Time 0 1 2 3Cash flow −1,000 x x x

Hence 1000 = x(ν + ν2 + ν3) = xa3 . Solving for x gives x = 381.05.

We can describe the repayment of the loan in the previous example as follows:

year loan outstanding repayment interest due capital repaid loan outstandingbefore repayment after repayment

1 l0 = 1000 x1 = 381.05 il0 = 70 x1 − il0 = 311.05 l1 = 688.952 l1 = 688.95 x2 = 381.05 il1 = 48.22 x2 − il1 = 332.83 l2 = 356.123 l2 = 356.12 x3 = 381.05 il2 = 24.93 x3 − il2 = 356.12 l3 = 0

4.2 General notation. Let

l0 = size of initial loani = effective rate of interest per time unit

For k = 1, 2, . . . , n, letlk = loan outstanding at time k immediately after repayment at time kxk = repayment at time kfk = capital repaid at time kbk = interest paid at time k

Hence

l0 = x1ν + x2ν2 + · · · + xnνn where ν =

11 + i


The loan schedule can be represented in a table:

year loan outstanding repayment interest due capital repaid loan outstandingbefore repayment after repayment

1 l0 x1 il0 x1 − il0 l1 = l0 − (x1 − il0)2 l1 x2 il1 x2 − il1 l2 = l1 − (x2 − il1)...

......

......

...k lk−1 xk ilk−1 xk − ilk−1 lk = lk−1 − (xk − ilk−1)...

......

......

...n ln−1 xn iln−1 xn − iln−1 ln = ln−1 − (xn − iln−1) = 0

4.3 Recurrence relations for the general case. We can use either a prospective loan calculation or a retro-spective loan calculation.

Retrospective loan calculation.At time t = 1,

the interest due is b1 = il0the capital repaid is f1 = x1 − il0leaving the outstanding loan of l1 = l0 − (x1 − il0) = l0(1 + i)− x1

And in general, for k = 1, 2, . . . , n we have:the interest due is bk = ilk−1the capital repaid is fk = xk − ilk−1leaving the outstanding loan of lk = lk−1(1 + i)− xk

These recurrence relations can be solved for lk as follows:l1 = l0(1 + i)− x1

l2 = l1(1 + i)− x2

= l0(1 + i)2 − x1(1 + i)− x2

and by induction for k = 1, 2 . . . , nlk = l0(1 + i)k − x1(1 + i)k−1 − x2(1 + i)k−2 − · · · − xk (4.3a)

In words, equation (4.3a) islk = (loan outstanding at time k immediately after the payment at time k)

= (value at time k of original loan)− (accumulated value at time k of repayments)

Prospective loan calculation.

At time t = n,the interest due is bn = iln−1the capital repaid is fn = ln−1hence the payment at time n must be xn = iln−1 + ln−1 = (1 + i)ln−1

And in general, for k = 1, 2, . . . , n− 1:the capital repaid is fk = lk−1 − lkhence the payment at time k must be xk = ilk−1 + (lk−1 − lk) = (1 + i)lk−1 − lkand hence lk−1 = νxk + νlk

These recurrence relations can be solved for lk as follows:ln−1 = νxnln−2 = νxn−1 + νln−1 = νxn−1 + ν2xn

and by induction for k = 0, 1 . . . , n− 1lk = νxk+1 + ν2xk+2 + · · · + νn−kxn (4.3b)

In words, equation (4.3b) islk = (loan outstanding at time k immediately after the payment at time k)

= value at time k of all future repayments


Example 4.3a. A loan is being repaid by 10 equal annual payments of £400. Suppose the effective annual interestrate is 12%. Find the loan outstanding immediately after the payment at the end of year 6.

Solution by the prospective method. The outstanding loan is the value at time 6 of all future repayments. 400(ν +ν2 + ν3 + ν4) = 400a4 ,0.12 = 1214.94Solution by the retrospective method. The outstanding loan is the value at time 6 of the original loan minus theaccumulated value at time 6 of the repayments. This is

(1.12)6l0 − 400s6 ,0.12 = (1.12)6 × 400a10 ,0.12 − 400s6 ,0.12 = 1214.94

Example 4.3b. A loan of £10,000 is to be repaid by 10 equal annual repayments of £1,400 plus a further finalrepayment one year after the last regular repayment. The effective annual interest rate is 12%. Find the size of thelast repayment.

Solution. Let ν = 1/1.12. The equation of value gives 10000 = 1400a10 ,0.12 + xν11. Hence x = 7269.08.

Example 4.3c. A loan of £9,000 is to be repaid by eight equal annual repayments of £1,400 plus a further finalrepayment one year after the last regular repayment. The effective annual interest rate is 12%. Find the outstandingloan after the eighth payment.

Solution. Using the retrospective method shows that the outstanding loan is the value at time 8 of the original loanminus the accumulated value at time 8 of the repayments. This is 9000× 1.128 − 1400s8 ,0.12 = 5064.10

4.4 A useful special case. Suppose a loan is repaid by n equal instalments of size x paid in arrears. Supposethe effective rate of interest is i per unit time.

Time 0 1 2 . . . n− 1 nCash flow, c −l0 x x . . . x x

Hence l0 = xan . The first payment of x consists of an interest payment of il0 = ixan = x(1− νn) and hencea capital repayment of x− il0 = xνn.After the kth payment, we have lk = xan−k by using the prospective method. Hence the (k + 1)th paymentconsists of an interest payment of ilk = ixan−k = x(1 − νn−k) and hence a capital repayment of x − ilk =xνn−k.So we have the general result: the kth repayment consists of an interest payment of x(1−νn−k+1) and a capitalrepayment of xνn−k+1 for k = 1, 2, . . . , n.

4.5 Repayments payable mthly. Now suppose the loan is repaid by nm instalments of x1/m at time 1/m,x2/m at time 2/m, . . . , xn at time nm/m = n.

Using the retrospective method, the loan outstanding after the kth repayment has been made is equal to(value at time k of original loan) − (accumulated value at time k of repayments)

Hence for k = 1, 2, . . . , nm1 we have

lk = (1 + i)k/ml0 − (1 + i)(k−1)/mx1/m − (1 + i)(k−2)/mx2/m − · · · − (1 + i)1/mx(k−1)/m − xk/m

Using the prospective method, the loan outstanding after the kth repayment has been made is equal to the valueat time k of all future repayments. Hence for k = 0, 1, . . . , nm− 1 we have

lk = ν1/mx(k+1)/m + ν2/mx(k+2)/m + · · · + ν(n−k)/mxn

At time k/m the loan outstanding is lk. Hence the interest incurred over the period (k/m, (k + 1)/m) is[(1 + i)1/m − 1

]lk =

i(m)lkm

Hence the capital repaid at time (k + 1)/m is

f(k+1)/m = x(k+1)/m −i(m)

mlk


Example 4.5a. Suppose a loan of size l0 is repaid by nm equal instalments of size x at times 1/m, 2/m, . . . ,nm/m =n. Suppose the interest rate charged is i% per annum effective. Find an expression for the capital repayment in thekth instalment.

Solution. Let ν = 1/(1 + i). Just after the payment of the (k− 1)th instalment, there are (nm− k + 1) instalments leftto pay and the loan outstanding, lk−1 is

lk−1 = x[ν1/m + ν2/m + · · · + ν(nm−k+1)/m

]Hence the capital repaid in the kth instalment is lk−1 − lk = xν(nm−k+1)/m for k = 1, 2, . . ., nm.

Example 4.5b. Suppose a loan of size l0 is either repaid in arrears by nm equal instalments of size x at times1/m, 2/m, . . . ,nm/m = n or it is repaid in advance by nm equal instalments of size y at times 0, 1/m, 2/m, . . . , (nm−1)/m.Suppose the interest rate charged is i% per annum effective. Show that x = (1 + i)1/my.

Solution. Let ν = 1/(1 + i). Then use

l0 = x[ν1/m + ν2/m + · · · + ν(nm)/m

]= y[1 + ν1/m + ν2/m + · · · + ν(nm−1)/m

]Example 4.5c.(a) The interest rate on a £5,000 loan is a nominal 12% per annum convertible 6-monthly. The borrower agrees torepay the loan by equal monthly instalments paid in arrears for 5 years. Find the amount of each repayment.

(b) Immediately after the 12th payment, the borrower renegotiates the loan: he repays a cash sum of £1,000 andagrees to repay the balance by equal monthly instalments over 2 years. The new interest rate is then fixed at anominal 11% per annum convertible 6 monthly. Find the amount of each repayment.

Solution.(a) Let x denote the size of each repayment. The monthly rate of interest is j = 1.061/6 − 1 = 0.009759. Then5,000 = xa60 ,j = x(1− ν60)/j where ν60 = 1/(1 + j)60 = 1/1.0610. Hence x = 110.49.(b) After the 12th payment, the loan outstanding is l12 = xa48 ,j . Define j1 by (1 + j1)6 = 1.055. Hence the newregular payment is y where ya24 ,j1

= l12 − 1,000 = xa48 ,j − 1,000. Hence y = 149.64.

4.6 Flat rate and APR Banks and other credit institutions sometimes quote a flat rate of interest. This iscalculated as follows: suppose a loan l0 is repaid by repayments over n years. Let xR denote the arithmeticsum of the total repayments. Then the flat rate of interest is

xR − l0nl0

=total interest paid

(length of loan in years)× (size of loan)Flat rates are really only useful for comparing loans of equal terms. The flat rate is always less than the trueeffective rate of interest charged on the loan.

In the UK, lenders are legally required to quote the effective rate of interest rounded to the lower 0.1%; this iscalled the annual percentage rate or APR.

4.7

Summary.• Suppose lk denotes the loan outstanding at time k immediately after the payment at time k.Retrospective loan calculation:

lk = (value at time k of original loan) − (accumulated value at time k of repayments)Prospective loan calculation:

lk = value at time k of all future repayments.

• Flat rate and APR.


5 Loans repaid by specified instalments5.1 A simple loan repaid by one instalment. Suppose a loan of size C bears interest at rate g% per annumpayable m times per year. The loan is repaid by a single payment of C after n years. Hence the amount gC ispaid in interest every year for n years.We have the following cash flow:

Time 0 1/m 2/m 3/m · · · nm−1/m nm/mCash flow −P gC/m gC/m gC/m · · · gC/m C + gC/m

Let i denote the yield obtained by purchasing this loan for the price P ; of course, this is also the effective rateof interest paid by the borrower. Then

P = gCa(m)n,i + Cνn = gC

1− νn

i(m) + Cνn =g

i(m) [C −K] +K (5.1a)

where K = Cνn is the value at time 0 of the redemption value C of the loan.It is easy to check that equation(5.1a) still holds if the loan is repaid after time k/m for any positive integer k.

5.2 A loan repaid by specified instalments. Suppose a loan of size C bears interest at rate g% p. a. payablem times per year. The loan is repaid by k instalments: the amount C1 at time n1/m, the amount C2 at time n2/m,. . . , and finally the amount Ck at time nk/m, where n1 < · · · < nk are positive integers and C = C1 + · · · +Ck.

Example 5.2a. A loan of £60,000 is issued with an interest rate of 8% p.a. payable quarterly in arrears. The loanwill be repaid by 3 equal annual instalments of £20,000 with the first instalment 5 years after the issue date. Find theprice P of the loan if the purchaser wishes to obtain a yield of 10% p.a.

Solution. Use units of £1,000. Initially, the annual interest is 4.8; hence each quarterly interest payment is 1.2. Attime 5 there is a capital repayment of 20; hence the outstanding loan is 40 and each quarterly interest payment is then0.8, and so on. Hence the cash flow is as follows:

Time 0 1/4 2/4 · · · 19/4 20/4 21/4 22/4 23/4 24/4 25/4 26/4 27/4 28/4Cash flow −P 1.2 1.2 · · · 1.2 21.2 0.8 0.8 0.8 20.8 0.4 0.4 0.4 20.4

Let x = 1/1.11/4, the quarterly discount factor. ThenP = 4.8a(4)

5 ,0.1 + 20x20 + 3.2x20a(4)1 ,0.1 + 20x24 + 1.6x24a(4)

1 ,0.1 + 20x28

=i

i(4)

[4.8a5 ,0.1 + 3.2x20a1 ,0.1 + 1.6x24a1 ,0.1

]+ 20

(x20 + x24 + x28)

= 21.58851281025 + 33.97106742687 = 55.55958023712 or £55,559.58.

Back to the general case. The value of the C1 instalment at time 0 is P1 = gi(m) [C1 −K1] + K1 where

K1 = C1νn1/m, and so on. Adding gives

P =g

i(m) [C −K] +K where K =k∑j=1

Cjνnj/m and C =

k∑j=1

Cj (5.2a)

Equation (5.2a) is called Makeham’s formula. Exercise 34 on page 68 extends this result to the situation wherethe loan is redeemable at 100α% for some α > 1.

Example 5.2b. Solve example 5.2a by using Makeham’s formula.

Solution. Use units of £1,000. In this solution, we decompose the cash flow as follows:Time 0 1/4 2/4 · · · 19/4 20/4 21/4 22/4 23/4 24/4 25/4 26/4 27/4 28/4

Cash flow −P1 0.4 0.4 · · · 0.4 20.4Cash flow −P2 0.4 0.4 · · · 0.4 0.4 0.4 0.4 0.4 20.4Cash flow −P3 0.4 0.4 · · · 0.4 0.4 0.4 0.4 0.4 0.4 0.4 0.4 0.4 20.4

Now P1 = gi(m)

[20− 20x20

]+ 20x20, P2 = g

i(m)

[20− 20x24

]+ 20x24, and P3 = g

i(m)

[20− 20x28

]+ 20x28. Adding

gives P = gi(m) [60−K] +K where K = 20(x20 + x24 + x28) = 20x20(1 + x4 + x8) = 33.97106742687. Hence

P =g

i(m) [C −K] +K =0.08

4(1.11/4 − 1)[60−K] +K = 55.55957881721 or £55,559.58.


Example 5.2c. A loan of £75,000 is issued with an interest rate of 8% p.a. payable quarterly in arrears. The loanwill be repaid by 15 equal annual instalments of £5,000 with the first instalment 5 years after the issue date. Find theprice P of the loan if the purchaser wishes to obtain a yield of 10% p.a.

Solution. Use units of £1,000. This could be solved from first principles as in example 5.2a but it would be tedious.Makeham’s formula can be used as follows: C = 75, K = 5(a19 ,0.1 − a4 ,0.1) = 25.975275. Hence

P =0.08

4(1.11/4 − 1)[75−K] +K = 66.63659797577 or £66,636.60.

5.3

Summary. Makeham’s formula.Suppose a loan of size C bears interest at rate g% p. a. payable m times per year. The loan is redeemedat 100α% for some α > 1. It is redeemed in k instalments: the amount C1 at time n1/m, the amount C2at time n2/m, . . . , and finally the amount Ck at time nk/m, where n1 < · · · < nk are positive integers.

Let i denote the yield and ν = 1/(1 + i). Let C ′ = C1 + · · · + Ck = αC and K =∑k

j=1 Cjνnj/m. Then

P =g

αi(m)

[C ′ −K

]+K


1. The annual rate of payment on a three year consumer credit contract is £2,000. The total charge for credit is £750.What is the flat rate of interest per annum? (Institute/Faculty of Actuaries Examinations, September 1999) [2]

2. A customer takes out a loan of £1,000 which is repaid by instalments of £703.84 p.a. payable annually in arrear fortwo years. Calculate the APR applicable to the loan.

(Institute/Faculty of Actuaries Examinations, May 1999. Specimen Examination.) [3]

3. A loan of £5,000 is repaid by an annuity payable annually in arrears for 12 years calculated at an effective rate ofinterest of 8% per annum. Find the interest element in the 5th payment.


4. A customer borrows £4,000 under a consumer credit loan. Repayments are calculated to give an APR of 15%.Instalments are paid monthly in arrears for 5 years. Calculate the flat rate of interest.


5. A credit company offers a loan of £5,000. The loan is to be repaid over a 4 year term by level monthly instalmentsof £130, payable in arrears. What is the APR for the transaction?


6. An individual purchases a car for £15,000. The purchaser takes out a loan for this amount that involves him making24 monthly payments in advance. The annual rate of interest is 12.36% per annum effective.

Calculate the flat rate of interest of the loan. (Institute/Faculty of Actuaries Examinations, September 2004) [5]

7. A 20 year loan of £100,000 is granted, to be repaid in monthly instalments paid in arrears on the basis of a rate ofinterest of 6% per annum convertible monthly. The rate of interest can be varied from time to time and the annualinstalment adjusted so that the capital outstanding at the time of the interest variation will be repaid by the remaininginstalments at the new rate of interest.

(i) Calculate the initial monthly instalment.(ii) After exactly 6 years, just after the payment then due, the interest rate is changed to 5.5% per annum convertible

monthly. Calculate the new monthly instalment.(Institute/Faculty of Actuaries Examinations, September 1999) [5]

8. An insurance company issues an annuity of £10,000 p.a. payable monthly in arrear for 20 years. The cost of theannuity is calculated using an effective rate of interest of 10% p.a.

(i) Calculate the interest component of the first instalment of the sixth year of the annuity.(ii) Calculate the total interest paid in the first 5 years.



9. A loan of £20,000 is being repaid by an annuity payable quarterly in arrears for 20 years. The annual amount of theannuity is calculated at an effective rate of interest of 10% per annum.

(i) Calculate the quarterly payment under the annuity.(ii) Calculate the capital and interest instalments in the 25th payment.


10. A bank makes a loan of £100,000 to an individual. The loan is to be repaid by level monthly instalments in arrearsover a period of 25 years. The instalments are such that the borrower pays interest at an effective rate of 5% perannum on the loan.

(i) Calculate the amount of the instalments.(ii) (a) Calculate the capital outstanding after 12 years, immediately after payment of the instalment then due.

(b) Determine the split of the 145th instalment between capital and interest.(Institute/Faculty of Actuaries Examinations, September 2003) [3+4=7]

11. Repaying a loan by a sinking fund. Suppose borrower A borrows £5,000 at an effective interest rate of 10% perannum. The borrower only pays the interest on the loan, but must repay all of the capital of the loan after 10 years.

(a) What is the annual interest payment?

(b) Suppose A pays a regular annual amount into a sinking fund which accumulates at an effective rate of 8%per annum. Find the size of the regular annual payment into the sinking fund in order to repay the loan after10 years.

(c) Compare the total annual payment of (a) and (b) with the regular payment that would be needed if the loan wasrepaid by the normal method of amortisation.

12. A company has borrowed £50,000 from a bank. The loan is to be repaid by level instalments, payable annually inarrears for 5 years from the date the loan is made. The annual repayments are calculated at an effective rate of interestof 10% per annum.

(i) Calculate the amount of the level annual payment and the total amount of interest which will be paid over the5 year term.

(ii) At the beginning of the third year, immediately after the second payment has been made, the company asks forthe loan to be rescheduled over a further 5 years from that date. The bank agrees to do this on condition that therate of interest is increased to an effective rate of 12% per annum for the term of the rescheduled payments andthat payments are made quarterly in arrears.(a) Calculate the amount of the new quarterly payment.(b) What is the interest content of the second quarterly instalment of the rescheduled loan repayments?


13. A company has borrowed £800,000 from a bank. The loan is to be repaid by level instalments payable annuallyin arrear for 10 years from the date the loan is made. The annual repayments are calculated at an effective rate ofinterest of 8% per annum.

(i) Calculate the amount of the level annual payment and the total amount of interest which will be paid over the10 year term.

(ii) At the beginning of the 8th year, immediately after the 7th payment has been made, the company asks for theterm of the loan to be extended by 2 years. The bank agrees to do this on condition that the rate of interestis increased to an effective rate of 12% per annum for the remainder of the term and that payments are madequarterly in arrear.(a) Calculate the amount of the new quarterly payment.(b) Calculate the capital and interest components of the first quarterly instalment of the revised loan repay-

ments. (Institute/Faculty of Actuaries Examinations, April 2007) [3+6=9]

14. A loan is to be repaid by an immediate annuity. The annuity starts at a rate of £100 p.a. and increases by £10 perannum. The annuity is paid for 20 years. Repayments are calculated using a rate of interest of 8% p.a. effective.

(i) Calculate the amount of the loan.

(ii) Construct a loan schedule showing the capital and interest elements in and the amount of loan outstanding afterthe 6th and 7th payments.

(iii) Find the capital and interest element of the last instalment.(Institute/Faculty of Actuaries Examinations, September 1998) [10]

15. An individual took out a loan of £100,000 to purchase a house on 1 January 1980. The loan is due to be repaid on1 January 2010 but the borrower can repay the loan early if he wishes. The borrower pays interest on the loan at a rateof 6% per annum convertible monthly, paid in arrears. The loan instalments only cover the interest on the loan. At thesame time, the borrower took out a 30 year investment policy, which was expected to repay the loan, and into whichmonthly premiums were paid in advance, at a rate of £1,060 per annum. The individual was told that premiums in


the investment policy were expected to earn a rate of return of 7% per annum effective. After 20 years, the individualwas informed that the premiums had only earned a rate of return of 4% per annum effective and that they wouldcontinue to do so for the final 10 years of the policy. The borrower agrees to increase his monthly payments into theinvestment policy to £5,000 per annum for the final 10 years.

(a) Calculate the amount to which the investment policy was expected to accumulate at the time it was taken out.

(b) Calculate the amount by which the investment policy would have fallen short of repaying the loan had extrapremiums not been paid for the final 10 years.

(c) Calculate the amount of money the individual will have, after using the proceeds of the investment policy torepay the loan, after allowing for the increase in premiums.

(d) Suggest another course of action the borrower could have taken which would have been of higher value to him,explaining why this higher value arises.

(e) Calculate the level annual instalment that the investor would have had to pay from outset if he had repaid theloan in equal instalments of interest and capital.


16. (i) Show that

(Ia)n =an − nνn

i(ii) (a) Calculate the present value, at a rate of interest of 3% per annum effective, of an annuity where 10 is paid

at the end of the first year, 12 is paid at the end of the second year, 14 is paid at the end of the third year,and so on, with payments increasing by 2 per annum until the payment stream ends at the end of 20 years.

(b) Calculate the capital outstanding in the above annuity at the end of the 15th year after the payment then duehas been received.

(c) Determine the interest and capital components in the 16th payment.(Institute/Faculty of Actuaries Examinations, September 2000) [12]

17. A bank makes a loan to be repaid in instalments annually in arrear. The first instalment is 50, the second 48 and so onwith the payments reducing by 2 per annum until the end of the 15th year after which there are no further payments.The rate of interest charged by the lender is 6% per annum effective.

(i) Calculate the amount of the loan.

(ii) Calculate the interest and capital components of the second payment.

(iii) Calculate the amount of the capital repaid in the instalment at the end of the 14th year.(Institute/Faculty of Actuaries Examinations, September 2005) [6+3+3=12]

18. A loan of £80,000 is repayable over 25 years by level monthly repayments in arrears of capital and interest. Therepayments are calculated using an effective rate of interest of 8% per annum.

Calculate(i) (a) The capital repaid in the first monthly instalment.

(b) The total amount of interest paid during the last 6 years of the loan.(c) The interest included in the final monthly payment.

(ii) Explain how your answers to (i)(b) would alter if, under the original terms of the loan, repayments had beenmade less frequently than monthly. (Institute/Faculty of Actuaries Examinations, April 2001) [12]

19. An individual borrows £100,000 from a bank using a 25 year fixed interest mortgage. The mortgage is repayableby monthly instalments paid in advance. The instalments are calculated using a rate of interest of 6% per annumeffective. After 10 years, the borrower has the option to repay any outstanding loan and take out a new loan, equalto the amount of the outstanding balance on the original loan, if interest rates on 15 year loans are less than 6% perannum effective at that time. The new loan will also be repaid by monthly instalments in advance.

(i) Calculate the amount of the monthly instalments for the original loan.

(ii) Calculate the interest and capital components of the 25th instalment.

(iii) The borrower takes advantage of the option to repay the loan after 10 years. The rate of interest on mortgagesof length 15 years has fallen to 2% per annum effective at that time. The first loan is repaid and a new loan istaken out, repayable over a 15 year period, for the same sum as the capital outstanding after 10 years on theoriginal loan.(a) Calculate the revised monthly instalment for the new loan.(b) Calculate the accumulated value of the reduction in instalments at a rate of interest of 2% per annum

effective, if the borrower exercises the option.(Institute/Faculty of Actuaries Examinations, September 2004) [3+4+6=13]


20. (i) A loan is repayable over 20 years by level instalments of £1,000 per annum made annually in arrear. Interest ischarged at the rate of 5% per annum effective for the first 10 years, increasing to 7% per annum effective for theremaining term.

Show that the amount of the original loan is £12,033.56. (Minor discrepancies due to rounding will not bepenalised.)

(ii) The following are the details from the loan schedule for year x, i.e. the year running from exact duration(x− 1) years to exact duration x years.

Loan outstanding at the Instalment paid at the end of the yearbeginning of the year Interest Capital

Year x £8,790.48 £439.52 £560.48Determine the value of x.

(iii) At the beginning of year 11, it is agreed that the increase in the rate of interest will not take place, so that therate remains at 5% per annum effective for the remainder of the term. The annual instalments will continue tobe payable at the same level so that there may be a reduced term and a reduced final instalment.(a) Calculate by how many years, if any, the repayment schedule is shortened.(b) Calculate the amount of the reduced final instalment.(c) Calculate the reduction in the total interest paid during the existence of the loan as a result of the interest

rate not increasing. (Institute/Faculty of Actuaries Examinations, April 2005) [2+4+7=13]

21. A loan is repayable by a decreasing annuity payable annually in arrears for 20 years. The repayment at the end ofthe first year is £6,000 and subsequent repayments reduce by £200 each year. The repayments were calculated usinga rate of interest of 9% per annum effective.

(i) Calculate the original amount of the loan.

(ii) Construct the schedule of repayments for years eight (after the seventh payment) and nine, showing the out-standing capital at the beginning of the year, and the interest element and the capital repayment in each year.

(iii) Immediately after the ninth payment of interest and capital, the interest rate on the outstanding loan is reducedto 7% per annum effective.Calculate the amount of the tenth payment if subsequent payments continue to reduce by £200 each year, andthe loan is to be repaid by the original date, i.e. 20 years from commencement.


22. A loan was taken out on 1 September 1998 and was repayable by the following increasing annuity.

The first repayment was made on 1 July 1999 and was £1,000. Thereafter, payments were made on 1 November,1 March and 1 July until 1 March 2004 inclusive. Each payment was 5% greater than its predecessor. The effectiverate of interest throughout the period was 6% per annum.

(i) Show that the amount of loan was £17,692 to the nearest pound.

(ii) Calculate the amount of capital repaid on 1 July 1999.

(iii) Calculate both the capital component and the interest component of the seventh repayment.(Institute/Faculty of Actuaries Examinations, April 2004) [5+3+7=15]

23. A loan of £10,000 was granted on 1 April 1994. The loan is repayable by an annuity payable monthly in arrears for15 years. The amount of the monthly repayment increases by £10 after 5 years and by a further £10 after 10 years,and was calculated on the basis of a nominal rate of interest of 8% per annum convertible quarterly.

(i) Calculate the initial amount of monthly repayment.

(ii) Calculate the amount of capital which was repaid on 1 May 1994.

(iii) Calculate the amount of loan which will remain outstanding after the monthly repayment due on 1 April 2002has been made.

(iv) The borrower requests that after the 1 April 2002 payment has been made, future repayments be for a fixedamount payable on each 1 April, the last repayment being on 1 April 2006. Calculate the amount of the revisedannual payment on this basis if the interest rate remains unchanged.


24. (i) Prove

(Ia)n =an − nνn

iA loan is repayable by an increasing annuity payable annually in arrears for 15 years. The repayment at the end ofthe first year is £3,000 and subsequent payments increase by £200 each year. The repayments were calculated usinga rate of interest of 8% per annum effective.(ii) Calculate the original amount of the loan.


(iii) Construct the capital/interest schedule for years 9 (after the eighth payment) and 10, showing the outstandingcapital at the beginning of the year, the interest element and the capital repayment.

(iv) Immediately after the tenth payment of interest and capital, the interest rate on the outstanding loan is reducedto 6% per annum effective.Calculate the amount of the eleventh payment if subsequent payments continue to increase by £200 each year,and the loan is to be repaid by the original date, i.e. 15 years from commencement.

(Institute/Faculty of Actuaries Examinations, April 2003) [3+3+6+5=17]

25. An actuarial student has taken out two loans.

Loan A: a 5 year car loan for £10,000 repayable by equal monthly instalments of capital and interest in arrear witha flat rate of interest of 10.715% per annum.

Loan B: a 5 year bank loan of £15,000 repayable by equal monthly instalments of capital and interest in arrear withan effective annual interest rate of 12% for the first 2 years and 10% thereafter.

The student has a monthly disposable income of £600 to pay the loan interest after all other living expenses havebeen paid.

Freeloans is a company which offers loans at a constant effective interest rate for all terms between 3 years and10 years. After 2 years, the student is approached by a representative of Freeloans who offers the student a 10 yearloan on the capital outstanding which is repayable by equal monthly instalments of capital and interest in arrear. Thisnew loan is used to pay off the original loans and will have repayments equal to half the original repayments.

(i) Calculate the final disposable income (surplus or deficit) each month after the loan payments have been made.

(ii) Calculate the capital repaid in the first month of the third year assuming that the student carries on with theoriginal arrangements.

(iii) Estimate the capital repaid in the first month of the third year assuming that the student has taken out the newloan.

(iv) Suggest, with reasons, a more appropriate strategy for the student.(Institute/Faculty of Actuaries Examinations, April 2006) [5+5+5+2=17]

26. A mortgage company offers the following two deals to customers for twenty-five year mortgages.

Product A. A mortgage of £100,000 is offered with level repayments of £7,095.25 annually in arrear. There are noarrangement or exit fees.

Product B. A mortgage of £100,000 is offered whereby a monthly payment in advance is calculated such that thecustomer pays an effective rate of return of 4% per annum ignoring arrangement and exit fees. In addition, thecustomer also has to pay an arrangement fee of £6,000 at the beginning of the mortgage and an exit fee of £5,000 atthe end of the twenty-five year term of the mortgage.

Compare the annual effective rates of return paid by the customers on the two products.(Institute/Faculty of Actuaries Examinations, April 2008) [8]

27. A government is holding an inquiry into the provision of loans by banks to consumers at high rates of interest. Theloans are typically of short duration and to high risk consumers. Repayments are collected in person by representa-tives of the bank making the loan. Campaigners on behalf of the consumers and campaigners on behalf of the banksgranting the loans are disputing one particular type of loan. The initial loans are for £2,000. Repayments are madeat an annual rate of £2,400 payable monthly in advance for 2 years.

The consumers’ association case. The consumers’ association asserts that, on this particular type of loan, consumerswho make all their repayments pay interest at an annual effective rate of over 200%.

The banks’ case. The banks state that, on the same loans, 40% of the consumers default on all their remainingpayments after exactly 12 payments have been made. Furthermore, half of the consumers who have not defaultedafter 12 payments default on all their remaining payments after exactly 18 payments have been made. The banksalso argue that it costs 30% of each monthly repayment to collect the payment. These costs are still incurred even ifthe payment is not made by the consumer. Furthermore, with inflation of 2.5% per annum, the banks therefore assertthat the real rate of interest that the lender obtains on the loan is less than 1.463% per annum effective.

(i) (a) Calculate the flat rate of interest paid by the consumer on the loan described above.(b) State why the flat rate of interest is not a good measure of the cost of borrowing to the consumer.

(ii) Determine, for each of the cases above, whether the assertion is correct.(Institute/Faculty of Actuaries Examinations, September 2007) [4+10=14]

28. A loan is repayable by annual instalments in arrear for 20 years. The initial instalment is £5,000 with each instalmentdecreasing by £200. The effective rate of interest over the period of the loan is 4% per annum.

(i) Calculate the amount of the original loan.(ii) Calculate the capital repayment in the 12th instalment.


After the 12th instalment is paid, the borrower and lender agree to a restructuring of the debt.The £200 reduction per year will no longer continue. Instead, future instalments will remain at the level of the 12th

instalment and the remaining term of the debt will be shortened. The final payment will then be a reduced amountwhich will clear the debt.(iii) (a) Calculate the remaining term of the revised loan.

(b) Calculate the amount of the final reduced payment.(c) Calculate the total interest paid during the term of the loan.


29. A company has borrowed £500,000 from a bank. The loan is to be repaid by level instalments, payable annuallyin arrear for ten years from the date the loan is made. The annual instalments are calculated at an effective rate ofinterest of 9% per annum.

(i) Calculate(a) the amount of the level annual instalments;(b) the total amount of interest which will be paid over the ten year term.

At the beginning of the eighth year, immediately after the seventh instalment has been made, the company asks forthe loan to be rescheduled over a further four years from that date. The bank agrees to do this on condition that therate of interest is increased to an effective rate of 12% per annum for the term of the rescheduled instalments and thatrepayments are made quarterly in arrear.(ii) (a) Calculate the amount of the new quarterly instalment

(b) Calculate the interest content of the second quarterly instalment of the rescheduled repayments.(Institute/Faculty of Actuaries Examinations, April 2012) [3+5=8]

30. A bank makes a loan to be repaid by instalments paid annually in arrear. The first instalment is £400, the second is£380 with the payments reducing by £20 per annum until the end of the 15th year, after which there are no furtherrepayments. The rate of interest charged is 4% per annum effective.

(i) Calculate the amount of the loan.(ii) Calculate the capital and interest components of the first payment.

At the beginning of the ninth year, the borrower can no longer make the scheduled repayments. The bank agrees toreduce the capital by 50% of the loan outstanding after the eighth repayment. The bank requires that the remainingcapital is repaid by a 10-year annuity paid annually in arrear, increasing by £2 per annum. The bank changes the rateof interest to 8% per annum effective.(iii) Calculate the first repayment under the revised loan.


31. A loan is to be repaid by an increasing annuity. The first repayment will be £200 and the repayments will increaseby £100 per annum. Repayments will be made annually in arrear for 10 years. The repayments are calculated usinga rate of interest of 6% per annum effective.

(i) Calculate the amount of the loan.(ii) (a) Calculate the interest component of the seventh repayment.

(b) Calculate the capital component of the seventh repayment.(iii) Immediately after the seventh repayment, the borrower asks to have the original term of the loan extended to

15 years and wishes to repay the outstanding loan using level annual repayments. The lender agrees but changesthe interest rate at the time of the alteration to 8% per annum effective. Calculate the revised annual repayment.


32. An individual takes out a 25-year bank loan of £300,000 to purchase a house. The individual agrees to pay only theinterest payments, monthly in arrear, for the first 15 years whereupon he repays half of the capital as a lump sum. Hethen pays only the interest for the remaining 10 years, quarterly in arrear, and repays the other half of the capital asa lump sum at the end of the term.

(i) Calculate the total amount of interest paid by the individual, assuming an effective rate of interest of 81/2% p.a.(ii) The individual believes that he can earn a nominal rate of interest convertible half-yearly of 9% per annum from

a separate savings account. Calculate the level contribution he must make monthly in advance to the savingsaccount in order to repay half the capital after 15 years.

(iii) The individual made the monthly contributions calculated in (ii) to the savings account. However, over the first15 years, the effective rate of return earned on the savings account was 10% per annum.The individual used the proceeds at that time to repay as much of the loan as possible and then decided to repaythe remainder of the loan by level instalments of interest and capital. After the first 15 years, the effective rateof interest changed to 7% per annum.Calculate the level payment he must make, payable monthly in arrear, to repay the loan over the final 10 yearsof the loan. (Institute/Faculty of Actuaries Examinations, September 2008) [5+4+5=14]


33. On 1 January 2016, a student plans to take out a 5-year bank loan for £30,000 that will be repayable by instalmentsat the end of each month. Under this repayment schedule, the instalment at the end of January 2016 will be X , theinstalment at the end of February 2016 will be 2X , and so on, until the final instalment at the end of December 2020will be 60X . The bank charges a rate of interest of 15% per annum convertible monthly.

(i) Prove that

(Ia)n =an − nνn

i(ii) Show that X = £26.62.

The student is concerned that she will not be able to afford the later repayments and so she suggests a revisedrepayment schedule. The student would borrow £30,000 on 1 January 2016 as before. She would now repay theloan by 60 level monthly instalments of 36X = £958.32 but the first repayment would not be made until the end ofJanuary 2019 and hence the final instalment is paid at the end of December 2023.(iii) Calculate the APR on the revised loan schedule and hence determine whether you believe the bank should

accept the student’s suggestion.(iv) Explain the difference in the total repayments made under the two arrangements.


34. An extension of Makeham’s formula in section 5 on page 61. A loan of nominal amount C bears interest at a rateof g% per annum payable m times per year. The loan is redeemable at 100α% after n years (this means there is apayment of αC after n years). Derive Makeham’s formula for this situation.The extension to k instalments is straightforward—this is a generalisation of equation (5.2a) .

35. A loan of £100,000 incurs interest at a rate of 10% per annum payable half-yearly. The loan is redeemable at 105%in five equal instalments after 4, 8, 12, 16 and 20 years. An investor who pays income tax at the rate of 20% buys theentire loan at issue for a price which gives a net yield of 8% p.a. effective.Find the price paid.

CHAPTER 4

Basic Financial Instruments

1 Markets, interest rates and financial instruments1.1 Markets. The term primary market refers to the original issue of a security, whilst the term secondarymarket refers to the market where the security is bought and sold after it is issued. The term market does notimply a physical marketplace—dealings may take place over the telephone or by computer.

The term money market refers to the market in short-term financial instruments which are based on an interestrate. Short-term usually means up to one year.The term capital market refers to all long-term financial instruments. These are predominantly bonds issuedby governments (domestic and foreign), government organisations, international bodies and companies withhigh credit status. Bond calculations are considered in the next chapter.The foreign exchange (or FOREX or FX ) market is the market where currencies are bought and sold.

This is not an exhaustive list of markets—for example, there are various commodities markets, the equitymarket, and so on. Markets are not independent—-their behaviour is linked by the return an investor canobtain.

1.2 Negotiable and bearer securities. A security is negotiable if it can be bought and sold in a secondarymarket; otherwise it is non-negotiable. A negotiable security is close to holding money.If a security is a bearer security then any payment is made to whoever is holding the security at the time. If asecurity is registered then the beneficial owner is the person recorded centrally as the owner; this central recordof ownership must be changed when the security is sold.

1.3 Money market financial instruments. The term money market refers to all short-term financial instru-ments. Short-term usually means up to one year.

Some of the main instruments in the money market are1

• Treasury bill or T-bill. Borrowing by government. A negotiable bearer security. Usually issued at a discount.Bills issued by the government and denominated in euros are called Euro bills.• Time deposit. Borrowing by a bank. Non-negotiable.• Certificate of deposit or CD. Borrowing by a bank. Negotiable and usually a bearer security. Usually carriesa coupon.• Commercial paper or CP. Borrowing by companies. A negotiable bearer security. Usually issued at adiscount.• Bill of exchange. Borrowing by companies. Used as an IOU between companies in order to finance trade.Negotiable.• Repurchase agreement or repo. Used to borrow short-term against a long term instrument. Non-negotiable.The difference between the purchase and repurchase prices implies the yield.• Futures contract. A deal to buy or sell some financial instrument at a later date.

In the UK, the term gilt or gilt-edged security denotes a UK Government liability in sterling, issued by HMTreasury and listed on the London Stock Exchange. The UK Debt Management Office (www.dmo.gov.uk) isthe government agency responsible for managing the government debt. It publishes several authoritative anduseful leaflets such as UK Government Securities: A Guide to Gilts, A Private Investors Guide to Gilts andFormulae for Calculating Gilt Prices from Yields. These can be downloaded from their web site—just choosethe Publications link. A table of current yields is also available on the web site.

The money market fulfils several functions:• One bank may have a temporary shortage and another may have a temporary surplus of money.

1 There are further notes about some of these financial instruments later in this chapter.



• Every bank will want to hold a proportion of its assets in a form in which it can quickly get at the money.• Companies, local authorities, and other bodies may have a temporary surplus or shortage of money.• The money market forms a bridge between the government and the private sector. If the government sellsT-bills then money is taken out of the market; if an eligible∗ bank is short of money it can get a loan fromthe government by selling a financial instrument to the Bank of England. These instruments include billsof exchange, gilt repos and Euro bills. The rate of interest at which the Bank is prepared to lend to thecommercial banks is called the Official Dealing Rate or the repo rate and is fixed by the infamous MonetaryPolicy Committee which meets every month. Its decisions are widely reported in the press.

1.4 Interest rates in the money market. Commercial banks have a bank base rate which is determined bythe Bank of England’s repo rate. Deposit rates and borrowing rates for individuals are calculated from thesebank base rates.

The London Inter-Bank Offered Rate or LIBOR is the rate for wholesale funds—it is the rate at which a bankwill lend money to another bank of top credit worthiness. The London Inter-Bank Bid Rate or LIBID is therate at which a bank bids for money from another bank. LIMEAN is the average of LIBOR and LIBID.Clearly, LIBOR > LIBID. LIBOR is a constantly changing measure of the cost of a large amount of money toa bank—it changes throughout the day!

The money market is linked to other markets through arbitrage mechanisms. For example, it is linked to thecapital market by the ability to create long-term instruments from a series of short-term instruments.

2 Fixed interest government borrowings2.1 Bonds. Governments (or government bodies) can raise money by issuing bonds. In most developedeconomies, government bonds form the largest and most liquid section of the bond market. They also constitutethe most secure long-term investment available. In the UK, bonds guaranteed by the government are calledgilt-edged securities or gilts. (Note that Treasury bills are for less than one year and are sold at a discount.)

In general, a borrower who issues a bond agrees to pay interest at a specified rate until a specified date, calledthe maturity date or redemption date, and at that time pays a fixed sum called the redemption value. The priceof the bond at issue is called the issue price.

The interest rate on a bond is called the coupon rate. This rate is often quoted as a nominal rate convertiblesemi-annually and is applied to the face or par value of the bond. The face and redemption values are often thesame.

Letf = face or par value of the bondr = the coupon rate (the effective coupon rate per interest period)C = the redemption value of the bondn = the number of interest periods until the redemption date

P = current price of the bondi = the yield to maturity (this is the same as the internal rate of return)

The values of f , r, C and the redemption date are specified by the terms of the bond and are fixed throughoutthe lifetime of the bond.

The values of P and i vary throughout the lifetime of the bond. As the price of a bond rises, the yield falls.Also, the price of a bond (and hence its yield) depends on the prevailing interest rates in the market.

If f = C then the bond is redeemable at par; if f > C then the bond is redeemable below par or is redeemableat a discount; if f < C then the bond is redeemable above par or is redeemable at a premium.

When a bond is sold between two coupon dates, the seller will be entitled to some accrued interest. This isexplained in section 1.8 of chapter 5 on page 85.

∗ The list of eligible banks can be found on the web site of the Bank of England.

4 Basic Financial Instruments Sep 27, 2016(9:51) Section 3 Page 71

2.2 Variations. Some bonds have variable redemption dates—the actual redemption date is chosen by theborrower (or occasionally the lender) at any interest date within a certain period or at any interest date after acertain date. In the latter case, the bond has no final redemption date and is said to be undated.

Some banks allow the interest and redemption proceeds of a bond to be bought and sold separately: this createszero coupon bonds and bonds with zero redemption value.

The real (or inflation adjusted) income stream from a bond will be volatile. Hence, governments sometimesissue bonds such that the coupon rate and redemption value are linked to an inflation index. Clearly this infla-tion indexation requires a specified time lag between the publication of the index figure and the recalculationof the coupon and redemption value. (So this means there is no inflation protection during the lag period.)

2.3 Government bills. These are for short term (less than one year) borrowing by government. They arealso called Treasury bills or T-bills and are negotiable bearer securities.

Government bills are usually issued at a discount and redeemed at par with no coupon. They are often used asa benchmark risk-free short-term investment.

Bills issued by the government and denominated in euros are called Euro bills.

2.4 Strips. There are no US or UK government zero-coupon bonds. However, investment banks may decom-pose a government bond into strips which stands for separately traded and registered interest and principalsecurities. Thus a bank may decompose a 3-year bond which has a semi-annual coupon into 7 components: the6 coupon payments and the final repayment of the principal. Each one of these 7 components is then equivalentto a zero-coupon bond and is traded separately.

3 Other fixed interest borrowings

3.1 Debentures and unsecured loan stocks. Debentures (or corporate bonds) form part of the loan capitalor long term borrowing of a company. They are usually secured against specific assets of the company.

Debentures are clearly not as secure as government bonds and so are less marketable and investors will expecta higher yield.

Unsecured loan stocks issued by a company will provide higher yields then debentures issued by the samecompany—this allows for the higher risk.

3.2 Foreign bonds and eurobonds. A samurai bond is a yen bond issued in Japan by a foreign (i.e. non-Japanese) organisation; a yankee bond is a dollar bond issued in the USA by a foreign (i.e. non-US) organisa-tion and a bulldog bond is a bond issued in sterling in the UK by a foreign (i.e. non-UK) organisation. Theseare examples of foreign bonds—bonds which are issued in the local currency but issued by a foreign borrower;the issue and secondary market is under the control of the local market authority.

Deposits of dollars in a European bank are called eurodollars. Deposits of French francs in a British bank arecalled eurofrancs. The prefix euro just indicates that the currency is being deposited outside of its countryof origin and the general term is eurocurrency. In order to make eurocurrency lending more marketable,eurobonds were developed.

Eurobonds are bonds issued by large companies, syndicates of banks or governments. They are usually unse-cured and interest is usually paid once per year. Thus a eurobond allows a company to borrow US dollars in theSwiss financial market, etc. In general, eurobonds are issued in a currency other than the issuer’s own currencyand in a country other than the issuer’s home country. Eurobonds provide medium or long-term borrowing freefrom the controls of any government. They are usually negotiable bearer securities with an active secondarymarket. Yields are typically less than the yields from conventional bonds from the same issuer with the sameterms. The features of eurobonds can vary—some examples are given in How to Read the Financial Pages byMichael Brett. The terminology is confusing: a “euro bond”, as opposed to a eurobond, is a bond denominatedin euros which could be issued domestically in the euro zone or internationally.


3.3 Certificates of deposit. These certificates are issued by banks and building societies in return for adeposit. They usually last from 28 days to 6 months and interest is paid on maturity. They are negotiable andusually a bearer security.

There is an active secondary market and marketability depends on the status and credit rating of the issuingbank. This marketability gives the investor flexibility: the CD can be sold if the investor needs the cash returnedsooner than expected and if general interest rates fall, then the CD can be sold for a capital profit. Because ofthis flexibility, the yield on a CD will typically be less than the yield on a deposit with the same bank and forthe same term.

3.4 CD calculations: CD with a single coupon. Consider a CD which pays a single coupon after d days.Let

f = face value of the CDic = the coupon rated = number of days interest the CD earnsdT = 360 for ACT/360 or 365 for ACT/365

The maturity proceeds are:

mp = f ×(

1 + icd

dT

)Clearly, ic is also the yield on the CD when it is issued.

Now suppose someone purchases the CD in the secondary market at time da for the price pa and sells it at timedb for the price pb where 0 < da < db < d. Let ia and ib denote the corresponding yields at the times da anddb.

......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

........................

........

........................ ............ ............

0issue

dapurchase(pa)

dbsale(pb)

dmaturity(mp)

time

The yield on this investment is (pbpa− 1)

dTdb − da

(3.4a)

Now

ia =(mp

pa− 1)

dTd− da

with a similar formula for ib. Substituting into equation (3.4a) gives the yield on the investment to be(1 + ia d−da

dT

1 + ib d−dbdT

− 1

)dT

db − da(3.4b)

Example 3.4a. Suppose a CD is issued with a face value of £500,000 and a coupon of 6% for 90 days.(a) After 30 days, its yield has fallen to 5.75%. What is its price?(b) After a further 30 days its yield has risen back to 6%. What is the rate of return for holding this CD for the 30days: day 30 to day 60. (Assume ACT/365.)

(a) The maturity proceeds are:

mp = f ×(

1 + icd

dT

)= 500,000

(1 + 0.06

90365

)= 507,397.26

Let pa denote the price after 30 days. Then(mp

pa− 1)

36560

= 0.0575 and so pa =mp

1 + 0.0575× 60/365= 502,646.22

(b) Using equation (3.4b) gives(365 + 0.0575× 60

365 + 0.06× 30− 1)

36530

= 0.05473 or 5.473%.


3.5 Repurchase agreements or repos. These are used to borrow short-term against a long term instrument.They are non-negotiable and the difference between the purchase and repurchase prices implies the yield.

Repos provide liquidity in the gilts market. The seller sells a gilt but agrees to buy it back at a specified laterdate at an agreed price. So the seller is really arranging a short-term loan with the gilt as collateral. If thespecified later date is the next day then it is called an overnight repo; otherwise it is called a term repo.

A reverse repurchase agreement or reverse repo is the same agreement but viewed from the perspective of theparty who loans the money and takes the gilt as security (the “buyer” or “investor”).

In summary, a repo is a purchase or sale of a security at the present time together with an opposite transactionlater. It makes possible the loan of cash or the loan of a specific security. Repos are used extensively byfinancial institutions to finance positions and they provide a link between the bond market and the moneymarket.

4 Investments with uncertain returns

4.1 Ordinary shares or equities. These entitle the holder to receive a share in the net profits of a company—after any interest on loans and fixed interest stocks (such as preference shares) has been paid. This payout iscalled the dividend. It is paid at the discretion of the directors, who may decide to retain profits which arethen used as reserves or to finance the company’s activities. Shareholders are the owners of a company and, inprinciple,2 can vote at the company’s Annual General Meeting.

The return on ordinary shares is made up of dividends and any increase or decrease in the market price. Theexpected return is higher than for most other asset classes—but the risks of loss are also higher3. Typically, theincrease in the price of an ordinary share is larger than the dividend received.

If a share is bought ex-dividend or xd then the seller, and not the buyer, receives the next dividend payment.Typically the xd date is about one month before the dividend payment date. The usual procedure with fixedinterest stocks is different—the buyer of a fixed interest stock usually compensates the seller for accruedinterest.

4.2 Preference shares. These are rarely used today. Like equities, they offer a stream of dividend income.However, preference shares usually pay a fixed dividend. They rank above ordinary shareholders for paymentof dividends and in winding-up but do not normally have voting rights unless the dividend is in arrears.

If an issue of preference shares has the abbreviation cum. in the title and the company cannot afford to pay thedividend, then the dividends are carried forward and paid when the company is able to do so. The abbreviationcum. stands for cumulative.

If the abbreviation red. occurs in the name of a preference share, then the shares are redeemable and will berepaid at some specified future date.

The return on preference shares is made nusually less than that on ordinary shares to reflect the lower risk.

4.3 Convertibles. Companies sometimes raise money by issuing convertible loan stock. This will have astated annual interest payment and is usually unsecured.

Convertible loan stock can be converted into ordinary shares at a fixed price at a fixed date in the future (or atone of a series of dates at the option of the holder). If not converted, it usually reverts to simple unsecured loanstock.

Convertible loan stock behaves like a cross between fixed interest stock and ordinary shares. As the date ofconversion approaches, it behaves more like the security into which it converts.

2 Because private investors usually hold their shares in a nominee account at a stockbroker, voting is often impossible.3 If a company fails and goes bankrupt, then the order in which debts are paid is: loan stock, then preference shares, then

ordinary shares. Hence, loan stock has lower risk than preference shares; preference shares have lower risk than ordinaryshares.


Convertible loan stock usually provides higher income than the dividend on ordinary shares and lower incomethan the dividend on preference shares or the interest paid on conventional loan stock. It combines the lowerrisk of a debt security with the potential large gains of an equity.

For the company, the main advantage of convertible loan stock is that the interest rate will be less than that onordinary unsecured loan stock. Investors accept the lower rate because of the possibility of capital gains fromthe rise in share prices. (The market value of convertibles will tend to follow the corresponding share price.)Also, if the company had issued ordinary shares, then its earnings and dividends would have been immediatelydiluted over a larger share capital base.

4.4 Property. Many different types of property available for investment: offices, shops, factories, etc. Thereturns from investing in property arise from the rental income and any capital appreciation achieved on itssale. Usually, both rental income and capital values are assumed to increase in line with inflation over the longterm. However, capital values can fluctuate substantially in the short term.

Rental terms are fixed in lease agreements. Typically rents increase every 3 to 5 years.

Note the following points about property investments:• less flexible than investment in equities or bonds• each property is unique and so can be difficult to value• buying and selling expenses are high• rental income is reduced by maintenance charges• Valuation is approximate and expensive because experienced surveyor must be employed• it may not be possible to rent the property and so rental income will fall

Because of the above drawbacks, the yield from property will normally be higher than that for equities.

The running yield of an investment is the annual income from the investment divided by its current marketprice; it ignores any capital growth. Hence the running yield of an equity is the dividend yield and the runningyield of a property is the rental yield. So ranking investments highest to lowest by running yield normally givesthe order: bonds, then property then equities.

5 DerivativesA derivative is a financial instrument with a value which is dependent on the value of some other financialasset. The purpose of the derivatives market is to redistribute risk. They also enable speculators to increasetheir exposure for a relatively modest outlay.

There are two main types of people using derivatives: those who want to hedge or guard against a risk and thetraders or speculators who are prepared to accept a high risk in return for the possibility of a high gain.

Types of risk include market risk and credit risk. Market risk is the risk that market conditions changeadversely—it is the risk that market conditions will change in such a way that the present value of the out-goings specified under an agreement increases. Credit risk is the risk that the other party to an agreement willdefault on its payments.

The main exchange-traded derivatives are futures and options. Forward contracts and swaps are over thecounter instruments.

5.1 Long and short positions.A long position in a security such as a bond (or equivalently to be long in a security) means the investor ownsthe security and will profit if the price of the security rises. This is the conventional position of an investorin the security. Short selling or shorting or going short in a security means selling a security that has beenborrowed from a third party with the intention of repurchasing it in the market at a later date and returning itto the third party. Hence a short seller makes a profit if the price of the security falls.

Similarly the holder of a long position in a derivative will profit if the price of the derivative rises whilst theholder of a short position in a derivative makes a profit if the price of the derivative falls.


5.2 Forward contract.A forward contract is a legally binding contract to buy/sell an agreed quantity of an asset at an agreed price atan agreed time in the future. The contract is usually tailor-made between two financial institutions or betweena financial institution and a client. Thus forward contracts are over-the-counter or OTC. Such contracts are notnormally traded on an exchange.

Settlement of a forward contract occurs entirely at maturity and then the asset is normally delivered by theseller to the buyer. Forward contracts are subject to credit risk—the risk of default.

5.3 Futures contract.A futures contract is also a legally binding contract to buy/sell an agreed quantity of an asset at an agreedprice at an agreed time in the future. Futures contracts can be traded on an exchange—this implies that futurescontracts have standard sizes, delivery dates and, in the case of commodities, quality of the underlying asset.The underlying asset could be a financial asset (such as equities, bonds or currencies) or commodities (such asmetals, wool, live cattle). A futures contract based on a financial asset is called a financial future.

Suppose the contract specifies that A will buy the asset S from B at the price K at the future time T . ThenB is said to hold a short forward position and A is said to hold a long forward position. Therefore, a futurescontract says that the short side must deliver to the long side the asset S for the exchange delivery settlementprice (EDSP). Thus the long side profits if the price rises and the short side profits if the price falls.

On maturity, physical delivery of the underlying asset is often not made—often the short side closes the contractby entering into a reverse contract with the long side and a cash settlement is made.

Each party to a futures contract must deposit a sum of money called the margin with the clearing house.Margins act as cushions against possible future adverse price movements. When the contract is first struck, theinitial margin is deposited with the clearing house. Additional payments of variation margin are made daily toensure the credit risk is controlled.

Note that futures contracts are guaranteed by the exchange on which they are traded. Many futures markets aremore liquid than the spot market in the underlying asset: for example, there are many types of UK governmentbonds but only one UK bond futures contract. Futures markets enable a trader to speculate on price movementsfor only a small initial cash payment and they also enable a trader to take a short position if he believes a priceis going to fall—he would sell a future contract.

A note on the differences between forward contracts and futures contracts.• Forward contracts are tailor-made contracts between the two parties. Futures contracts are traded on an exchange—consequently, futures contracts are standardised (size of contract, delivery date, etc.)• Settlement of forward contracts occurs at maturity. With futures contracts, there is the process of marking to marketdescribed in chapter 5.• For most futures contracts, delivery of the underlying asset is not normally made. A cash flow is made to close outthe position. With forward contracts, delivery is usually carried out.

I understand forward contracts started in Chicago in the 1840s. Farmers in the Midwest shipped their grain to Chicagofor sale and distribution. Because all the grain was harvested at the same time, the price of grain dropped drastically atharvest time and then gradually increased as the stock was consumed. It made sense for farmers to agree to a forwardcontract in which they agreed to supply an agreed amount of grain at an agreed price at some time in the future.

Subsequently, hedgers and speculators wished to trade in these contracts. This led to the establishment of an exchangewhich laid down rules for contracts which could be traded and also checked the financial status of both parties to thecontract—this led to futures contracts. The explosion in the use of futures contracts occurred with the introduction offinancial futures. I believe that Chicago is still the second largest futures exchange in the world.For further information see the two books by J.C. Hull and M. Brett (page 290 onwards).

5.4 Some special cases of futures contracts.• The bond future. Government bond futures contracts are one of the most popular futures contracts.

A bond futures contract is an agreement to buy or sell a bond at a specified price at a specified time in thefuture. Some contracts are usually specified in terms of a notional bond and then there needs to be an agreedlist of bonds which are eligible for fulfilling the contract. The short side will then choose the bond from thelist which is cheapest to deliver. The price paid by the long side may have to be adjusted to allow for the fact


that the coupon on the bond which is actually delivered may not be equal to the coupon on the notional bondspecified in the contract. Most bond futures contracts are closed out prior to maturity and so actual deliverydoes not occur.

Suppose a fund manager has some UK gilts and expects a rise in interest rates. The rise in interest rates willlower the value of his gilts—so he can hedge against this by selling a bond futures contract. If the rise ininterest rate occurs, then the price of the gilts and the futures price of the bonds will fall. This will offset theloss in the value of his gilt-edged stock.

Example 5.4a. On January 10, 2001, the settlement price of the “US Treasury Bond $100,000; pts. 32nds of 100%”for March delivery is given as 123.06.

In finance, a basis point denotes 0.01%. So the expression “pts. 32nds of 100%” implies that 123.06 denotes 123basis points plus 6/32 which is 1236/32%.The listed price of 123.06 means 1236/32% of par, which is 1236/32× $100,000/100 = $123,187.50.

Hence, if the price of this contract rises one point to 124.06 then the long side gains $1,000 and the short side losesthe same amount. If the initial margin was $10,000 then the long side would have made 1,000/10,000×100% = 10%profit. Note that the value of the bond has only increased by 1/(1236/32)× 100% = 0.8%.

A basis point is always 0.01%; a tick is the minimum price change allowed in trading. The value of a tickvaries from instrument to instrument: it can be a basis point or 1/2 basis point or 1/4 basis point, etc.

• The stock index future. The contract provides for a transfer of assets underlying a stock index at a specifiedprice on a specified date. Settlement is by a cash transfer.

For example, there are futures based on the FTSE 100 index and these provide a way of betting on the move-ment of the market as a whole. Each FTSE futures contract is priced at £25 per index point and has one of4 possible delivery dates each year: March, June, September and December.

A bull will buy a FTSE futures contract: each rise of 1 point will give him a gain of £25. A bear will sell aFTSE futures contract: each fall of 1 point will give a profit of £25.

• The currency future. The contract requires the delivery of a specified amount of a given currency on aspecified date.

5.5 Special cases: forward rate agreement (FRA) and interest rate future.

• Forward rate agreement or FRA. An FRA is a forward contract on a short-term loan. It is an interest ratehedge which allows the borrower to fix the cost of money he intends to borrow in a few months’ time. Theborrower pays a fixed interest rate and in return receives a floating interest rate—this floating interest rate issome agreed reference rate such as LIBOR. An FRA is a forward contract and is an OTC equivalent of aninterest rate future—these are explained below.

For example, a 3 × 9 FRA (or 3v9 FRA) is a 3-month forward on a 6-month loan. The interest on the loanis set when the contract is arranged and the amount of the loan is notional—this means that no loan is everextended4. The only cash transfer is the difference in interest between using the agreed interest rate and thereference interest rate on the date the notional loan commences—this date is called the effective date.

Example 5.5a. Consider a 3 × 9 FRA for £1,000,000 with an FRA rate of 3.4%. Suppose the reference rate isLIBOR and the 6-month LIBOR on the effective date is 3.7%. Assume ACT/360 and the loan is for a period of180 days. Find how much the borrower receives from the lender on the effective date.

Solution. The borrower will receive

£1,000,000(0.037− 0.034)× 180/360

1 + 0.037× 180/360= £1472.75

Note that the amount £1,000,000× (0.037− 0.034)× 180/360 is the difference in interest payments due at the endof the loan. This quantity is divided by 1 + 0.037× 180/360 to bring it into the value at the effective date.

Because the loan is notional, an FRA is an off-balance sheet instrument.

4 A forward-forward contract is an agreement to borrow an agreed amount at some specified time in the future and repay itwith specified interest at some specified later date. For such a contract, the principal of the loan is paid.


• Interest rate future. An interest rate future is similar to an FRA but instead is a futures contract—hence itis traded on an exchange and there is the process of “marking to market” instead of a single cash payment.

The contracts are structured so that the price of the contract rises as the interest rate falls. Typically, the priceis stated as 100− x where x is the interest rate. For example, if x = 6.25 then the contract is priced at 93.75.

On expiry, the purchaser will have made a profit or a loss depending on the difference between the finalsettlement price and the original dealing price. The contract is settled by a cash transfer.

A borrower who wishes to hedge against a rise in interest rates should sell an interest rate future. A lender whowishes to hedge against a fall in the interest rate should buy an interest rate future.

Note that FRAs and futures move in opposite directions. A buyer of an FRA profits if the interest rate riseswhilst a buyer of an interest rate future profits if the interest rate falls. Thus a trader who sells an FRA will alsosell an interest rate future in order to cover his position.

Example 5.5b. On January 10, 2001, the price of the three month Euromark futures contract on LIFFE for DM 1million, points of 100% is quoted as 97.02 for March, 2001.

This means that a buyer of this contract on January 10, 2001 agrees to lend 1 million DM to the seller of the contractfor 3 months from March, 2001 at 100− 97.02 = 2.98%.

The phrase “points of 100%” explains the pricing unit. A basis point is 0.01%. Hence the smallest possible pricemovement is 0.01% over 3 months which gives 1,000,000 × 0.0001 × 0.25 = 25 DM (we use 0.25 because it is a3 month contract). This smallest possible price movement is called a tick.If the price changes to 97.23 or moves up 21 ticks, then this is equivalent to 21× 25 = 525 DM.

Example 5.5c. A dealer purchases a three month Euromark futures contract for DM 1 million at 96.29. He closesthe contract at 96.22. Find his loss.

Solution. His loss is DM1,000,000× 0.0007× 0.25 = DM175.

The price of an interest rate future is determined by the market. In the last example, the trader does not lendthe principal of DM1,000,000 at the implied interest rate of (100 − 96.22)% = 3.78%. He just makes a profitor loss depending on interest rate at the time of settlement.

An FRA settlement is always discounted but the settlement of an interest rate future is not usually discounted.

5.6 Currency swaps. This is an agreement to exchange a specified series of interest payments and a specifiedcapital sum in one currency for a specified series of interest payments and a capital sum in another currency.For example, a French company may require US dollars but can borrow French francs more cheaply thanUS dollars because the company is well-known in France and not well-known in the US. Conversely, anothercompany may require French francs but can borrow US dollars more cheaply. So if both companies borrow inthe cheapest way and arrange a currency swap then both will end up with cheaper funds. In practice, both willuse a bank which arranges many swaps of this type.

5.7 Interest rate derivatives. In an interest rate swap, one party pays a series of fixed payments for a fixedterm. In return, the other party pays a series of variable payments—these are the interest on the same (notional)deposit at a floating rate.

Swaps are priced so that the present value of the cash flows for the investor is slightly negative. This is theprofit margin for the issuer.

Both parties face both market and credit risk. Market makers will often hedge market risk by making anoffsetting agreement. The credit risk is not as great as that on a loan of the notional principal because the riskonly occurs if the swap has a negative value to the defaulting party.

Example 5.7a. CompanyA can borrow at a floating rate of LIBOR+0.1% or at a fixed rate of 8.0% p.a. CompanyBis not so creditworthy and can borrow at a floating rate of LIBOR + 0.8% or at a fixed rate of 9.5% p.a. Company Aprefers to borrow at a floating rate and company B prefers to borrow at a fixed rate.

Suppose company A borrows at a fixed rate and company B borrows at a floating rate and they swap at 8.3% p.a.The cost to company A is 8.0%− 8.3% + LIBOR = LIBOR− 0.3%.The cost to company B is LIBOR + 0.8% + 8.3%− LIBOR = 9.1%.Hence both sides gain by 0.4% p.a.


In practice, both companies would approach a bank to arrange an interest rate swap and the high-street clearingbanks sell interest rate swaps over-the-counter (OTC). For example, a company may wish to swap the intereston a floating-rate loan it has arranged for a fixed rate of interest. Then the bank assumes responsibility forpaying the floating rate and charges the company a fixed rate.

A company may wish to arrange a cap—this means that the bank will pay any interest above a specified agreedrate. A floor is an option which works the other way: the company receives a payment from the bank providedit agrees to pay a minimum rate even if the interest rate falls below this level. Arrangements that include a capand a floor are called a collar or cylinder.

All these interest rate derivatives act as hedges against interest rate movements. Of course, the bank also needsto hedge its risk by trying to match up the different swaps it issues, etc.

5.8 Options.A call option gives the right (but not the obligation) to buy a specified asset for a specified price (the strikeprice or exercise price) at a specified time in the future.A put option gives the right (but not the obligation) to sell a specified asset for a specified price (the strike priceor exercise price) at a specified time in the future. Note that purchasing a put option does not mean purchasingthe underlying asset. The holder of a put option has just purchased the right to sell the asset at a certain price atsome time in the future. If he wishes to exercise that right, then he must first purchase the asset (at the marketprice prevailing at the expiry date) and then sell it.

An American option means that the option can be exercised on any date before its expiry; a European optionmeans that the option can only be exercised on the expiry date.

Clearly an investor who is bullish will buy a call option, whilst an investor who is bearish will buy a put option.

Financial options are traded in standard units called contracts; for example, a contract for 100 shares. In theUK, financial options are traded on Liffe, the London International Financial Futures and Options Exchange.Thus the owner of a traded option may sell the option to someone else, exercise the option to buy (or sell) theunderlying asset at the fixed price, or abandon the option with no delivery of the underlying asset.

A call option is said to be in-the-money if the price of the underlying asset is above the exercise price and is saidto be out-of-the-money if the price of the underlying asset is below the exercise price. A put option is said to bein-the-money if the price of the underlying asset is below the exercise price and is said to be out-of-the-moneyif the price of the underlying asset is above the exercise price.

The plot on the left in figure (5.8a) shows the profit/loss at expiry for the holder of a call option. The holderbreaks even if the price of the asset at expiry equals the exercise price plus the premium paid. Theoretically, theprice of the asset could rise to any amount, and so the potential profit is unbounded above. The maximum lossis the premium paid—this occurs if the value of the asset at expiry is 0. The profit/loss diagram for the writerof a call option is the mirror image in the y-axis of this graph—this means that his potential loss is unlimited ifhe does not own the asset.

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−Premium −Premium

0 0

Profit/loss at expiry for a call option holder Profit/loss at expiry for a put option holder

Figure 5.8a.(PICTEX)

The plot on the right in figure (5.8a) is the profit/loss at expiry for the holder of a put option. The holder ofa put option cannot gain more than the exercise price less the premium paid—this occurs if the value of theunderlying asset drops to 0 at expiry. The holder of the put option will then exercise his right to sell the asset

4 Basic Financial Instruments Sep 27, 2016(9:51) Exercises 6 Page 79

at the exercise price. The holder of a put option cannot lose more than the premium paid—this occurs if thevalue of the underlying asset is above the exercise price. In that situation, the holder of the put option will notexercise his right to sell the asset and so he will just lose the premium he has paid.The profit/loss diagram for the writer of a put option is the mirror image in the y-axis of the previously describedgraph—this means that his maximum loss is the exercise price minus the premium.

5.9 Dangers of derivatives. Trading in futures can be more dangerous than trading in the underlying assetbecause of gearing or leverage. The initial margin will be small compared with the potential loss. Thus it ispossible to lose large amounts (and gain large amounts) with a relatively small initial outlay.

For purchasers of options, the maximum loss is limited to the initial outlay.


1. A CD (Certificate of Deposit) is issued for £1,000,000 on 12 March for 90 days with a 5% coupon. Hence it matureson 10 June.(a) What are the proceeds on maturity?(b) On 4 April, what should the secondary market price be in order that the yield is then 4.5%?(c) Assume the CD is purchased in the secondary market on 4 April for the price calculated in part (b). By 4 May,

the yield has dropped to 4%. What is the rate of return for holding the CD from 4 April to 4 May?(Assume ACT/365.)

2. (a) Distinguish between a future and an option.(b) Explain why convertibles have “option-like” characteristics.


3. Describe how cash flows are exchanged in an “interest rate swap”.(Institute/Faculty of Actuaries Examinations, September 2005) [2]

4. State the main differences between a preference share and an ordinary share.(Institute/Faculty of Actuaries Examinations, April 2003) [3]

5. Describe the features and risk characteristics of a “Government Bill”.(Institute/Faculty of Actuaries Examinations, September 2002) [4]

6. (i) Define the characteristics of a government index-linked bond.(ii) Explain why most index-linked securities carry some inflation risk, in practice.


7. A company has entered into an interest rate swap. Under the terms of the swap, the company makes fixed annualpayments equal to 6% of the principal of the swap. In return, the company receives annual interest payments on theprincipal based on the prevailing variable short-term interest rate which currently stands at 5.5% per annum.(a) Describe briefly the risks faced by a counterparty to an interest rate swap.(b) Explain which of the risks described in (a) are faced by the company.


8. State the characteristics of an equity investment.(Institute/Faculty of Actuaries Examinations, September 2007) [4]

9. Describe the characteristics of the following investments:(a) Eurobonds(b) Certificates of Deposit (Institute/Faculty of Actuaries Examinations, April 2008) [4]

10. Describe the characteristics of the cash flows that are paid and received in respect of(i) an index-linked security

(ii) an equity (Institute/Faculty of Actuaries Examinations, September 2013) [2+3=5]

11. Describe the characteristics of commercial property (i.e. commercial real estate) as an investment.(Institute/Faculty of Actuaries Examinations, September 2008) [5]

12. Explain why the running yield from property investments tends to be greater than that from equity investments.(Institute/Faculty of Actuaries Examinations, April 2015) [3]

CHAPTER 5

Bonds, Equities and Inflation

1 Bond calculations1.1 Notation. A borrower who issues a bond agrees to pay interest at a specified rate until a specified date,called the maturity date, and at that time pays a fixed sum called the redemption value.

The interest rate on a bond is called the coupon rate. This rate is often quoted as a nominal rate convertiblesemi-annually and is applied to the face or par value of the bond. The face and redemption values are often thesame. Let

f = face or par value of the bondr = the coupon rate per yearC = the redemption value of the bondn = the number of years until the redemption dateP = current price of the bondi = the yield to maturity (this is the same as the internal rate of return)

The values of f , r, C and the date of redemption are specified by the terms of the bond and are fixed throughoutthe lifetime of the bond.

The values of P and i vary throughout the lifetime of the bond. As the price of a bond rises, the yield falls.Also, the price of a bond and hence its yield depend on the prevailing interest rates in the market and the riskof default.

1.2 Finite redemption date.

Example 1.2a. Suppose a bond with face value £500,000 is redeemable at par after 4 years. The coupon rate is10% p.a. convertible semi-annually.(a) Find the price to obtain an effective yield of 10% p.a. (b) Find the price to obtain an effective yield of 15% p.a.

Solution. For this example, f = C = 500,000; r = 0.1, and n = 4. Each coupon payment is £25,000. The cash flowin thousands is as follows:

Time (years) 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0Cash flow (£1,000’s) −P 25 25 25 25 25 25 25 525

(a) The price P (in thousands) for a yield of i = 10% is given by

P =25

1.10.5 +251.1

+25

1.11.5 +25

1.12 +25

1.12.5 +25

1.13 +25

1.13.5 +5251.14

Let α = 1/√

1.1. Then

P = 25(α + α2 + · · · + α8) + 500α8 = 25α1− α8

1− α+ 500α8 =

11.14

[25

1.14 − 11.11/2 − 1

+ 500]

= 503.868

Alternatively, let i = 0.1 and ν = 1/(1 + i), then

P = 50a(2)4 ,i +

500(1 + i)4 = 50

i

i(2) a4 ,i + 500ν4 = 503.868

Hence the bond must be bought at a premium.Note that commonly used values of i/i(m), νn and an are provided in the tables.(b) For this case

P =1

1.154

[25

1.154 − 11.151/2 − 1

+ 500]

= 433.792

Alternatively, P = 50a(2)4 ,i + 500/(1 + i)4 = 50 i

i(2) a4 ,i + 500/(1 + i)4 = 433.792.Hence the bond is sold at a discount.



Here is the general result: suppose the coupon rate is nominal r% p.a. payable m times per year. Then theprice P in order to achieve an effective yield of i per annum is given by

P = fra(m)n,i +

C

(1 + i)n=fr

mamn,i′ +

C

(1 + i′)mn(1.2a)

where (1 + i′)m = 1 + i and i′ = i(m)

m is the effective interest rate per period. If the price rises, then the yieldfalls, and conversely.If P = f = C, then using the relation a(m)

n = (1 − νn)/i(m) and equation (1.2a) shows that i(m) = r. Thismeans that if the face value, redemption value and price are all the same, then the nominal yield convertiblemthly equals the coupon rate.If the income tax rate is t1, then equation (1.2a) becomes:

P = fr(1− t1)a(m)n,i +

C

(1 + i)n=g(1− t1)i(m)

[C − Cνn

]+ Cνn (1.2b)

where the right hand version of equation(1.2b) is called Makeham’s formula. The first term is the NPV of theinterest payments and the second term is the NPV of the redemption value, C.

Example 1.2b. Suppose a bond is issued for 15 years with a coupon rate of 6% per annum convertible 6-monthly.The bond is redeemable at 105%. (This means that C = f × 1.05.)(a) Find the price per unit nominal 1 so that the effective yield at maturity is 7% per annum.(b) Assume the income tax rate is 35%. Find the price per unit nominal so that the effective yield at maturity is7% p.a.

Solution. For this example, we have f = 100, C = 105, r = 0.06, i = 0.07 and n = 15.(a) The size of each coupon is 3. Hence

P = 6a(2)15 ,i +

105(1 + i)15 = 6

i

i(2) a15 ,i +105

(1 + i)15 ≈ 93.64

by using tables. Hence the bond should be issued at a discount. Part (b):

P = 6× 0.65× a(2)15 ,i +

105(1 + i)15 = 3.9

i

i(2) a15 ,i +105

1.0715 ≈ 74.19

1.3 Gross yield, redemption yield and flat yield.The terms gross interest yield, direct yield, flat yield, current yield or gross running yield all refer to 100r/P1,which is the ratio of 100r, the annual coupon per £100, to the current price P1 per unit nominal1. This is thesame as the annual coupon (fr) divided by the current price, P .The terms net interest yield and net running yield refer to the same quantity but after allowing for tax: hence itis 100(1− t1)r/P1, the ratio of the (after tax) annual coupon per £100 to the current price P1 per unit nominal.This is the same as (1− t1)fr/P , the annual coupon after tax divided by the current price, P .

Example 1.3a. The current price of a bond with an annual coupon of 10% is £103 per £100 nominal. Then thegross interest yield (also known as the flat yield or running yield) is 10/103 = 0.0971 or 9.71%.

The interest yields defined in the last paragraph are useful when a bond is undated and there are no redemptionproceeds. If there are redemption proceeds, then we should also take these into account when calculating theyield.The term gross redemption yield is the same as the yield to maturity defined in paragraph 1.1 and ignorestaxation. The term net redemption yield or net yield to redemption or net yield refers to the yield afterallowing for tax.

If the gross redemption yield on a bond is an effective 3% every six months, then the gross redemption yield is6.09% per annum (because 1.032 − 1 = 0.0609). Alternatively, it is described as a gross redemption yield of6% convertible 6-monthly.2

1 In the UK, per unit nominal means per £100.2 In Europe, the gross redemption yield is the same as the internal rate of return as defined in paragraph 1.1. However, in

the USA and the UK, the gross redemption yield of a bond with 6-monthly coupons is defined to be twice the effectiveyield for 6 months—what we have defined to be the gross redemption yield convertible 6-monthly. In actuarial questions,the distinction will be clarified by using phrases such as “a gross redemption yield of 6.09% per annum effective” or “agross redemption yield of 6% per annum convertible 6-monthly”. See Basic Bond Analysis by Joanna Place which canbe downloaded from http://www.bankofengland.co.uk/education/Pages/ccbs/handbooks

5 Bonds, Equities and Inflation Sep 27, 2016(9:51) Section 1 Page 83

Example 1.3b. Consider a US treasury bond which has 14 years to redemption and a coupon of 12% p.a. payable6-monthly. What is the price of the bond (per $100) if the gross redemption yield is 8% convertible 6-monthly.

Solution. The yield is 4% effective for 6 months.

Time 0 1/2 1 3/2 2 5/2 · · · 27/2 14Cash flow −P 6 6 6 6 6 · · · 6 106

So let ν = 1/1.042 and α = ν1/2 = 1/1.04. Hence

P = 6(α + α2 + α3 + · · · + α27 + α28) + 100α28 = 6α1− α28

1− α+ 100α28 = 133.326

In practice, the most common problem is finding the gross redemption yield when the current market price isgiven. Suppose £P denotes the price per £100 face value, there are n years to redemption and coupons are paidevery 6 months. Let i′ denote the effective yield per 6 months and let α = 1/(1 + i′). Then

P =fr

2(α + α2 + · · · + α2n) + 100α2n =

fr

2α

1− α2n

1− α+ 100α2n =

fr

21− α2n

i′+ 100α2n (1.3a)

Finding i′ may require successive approximation on a calculator. Equation (1.3a) impliesfr

2i′+

α2n

1− α2n (100− P ) = P and so 2i′ =fr

P+

100− PP

2i′

(1 + i′)2n − 1≈(fr +

100− Pn

)1P

(1.3b)

by using the approximation (1 + i′)2n ≈ 1 + 2ni′. This approximation for i′ is useful as a starting value whenfinding an approximate solution. It can be remembered as the gross redemption yield payable 6-monthly (2i′)approximately equals (coupon per year + capital gain per year)/price. This approximation will be greater thanthe exact value.

1.4 Perpetuity. Suppose a security is undated—this means that it has no final redemption date. Assume thecoupon is r% payable half-yearly. The cash flow is as follows:

Time 0 1/2 1 3/2 2 5/2 3 7/2 · · ·Cash flow −P fr/2 fr/2 fr/2 fr/2 fr/2 fr/2 fr/2 · · ·

Assume an income tax rate of t1. Then the price P to achieve an effective yield of i per annum is:

P = fr(1− t1)a(2)∞ =

fr(1− t1)i(2)

1.5 Effect of capital gains tax on bond prices. Suppose we have a bond with face value f , redemptionvalue C, coupon rate r% payable m times per year for n years. Let

g =fr

C=

annual couponredemption value

Then P (n, i), the price of a bond with an effective annual yield of i, is given by equation (1.2b):

P (n, i) = (1− t1)fra(m)n,i + Cνn (1.5a)

Using a(m)n,i = (1− νn)/i(m) gives

P (n, i) = (1− t1)gCa(m)n,i + C

[1− i(m)a(m)

n,i

](1.5b)

= C +[(1− t1)g − i(m)]Ca(m)

n,i (1.5c)

Capital gains tax is levied on the difference between the sale and purchase price of a security. It is levied onlywhen the security is sold. Suppose that capital gains tax is levied at the rate t2; this alters the price of the bondif the yield remains at i. Let the new price of the bond be P2(n, i). Equation (1.5c) shows that:

• If i(m) ≤ (1− t1)g, then P ≥ C. There is no capital gain and so P2(n, i) = P (n, i).• If i(m) > (1− t1)g, then P (n, i) < C. Hence there is a capital gain3. Hence

P2(n, i) = (1− t1)fra(m)n,i + Cνn − t2

[C − P2(n, i)

]νn

3 Aide memoire. If the coupon rate is very high (and hence g is large) then the bond will be in high demand; hence its pricewill be high and there will be no capital gain.


Hence

P2(n, i) =(1− t1)fra(m)

n,i + (1− t2)Cνn

1− t2νn

and so

P2(n, i) =

(1− t1)fra(m)

n,i + Cνn if i(m) ≤ (1− t1)g

(1− t1)fra(m)n,i + (1− t2)Cνn

1− t2νnif i(m) > (1− t1)g

Example 1.5a. A bond with face value £1,000 is redeemable at par after 10 years and has a coupon rate of 6%payable annually. The income tax rate is 40% and the capital gains tax rate is 30%. If the current price of the bondis £800, what is the net yield?

Solution. After allowing for income tax of £24 on each coupon and capital gains tax of £60, the cash flow is asfollows:

Time 0 1 2 3 . . . 9 10Cash flow −800 36 36 36 . . . 36 36 + 940

Hence the equation for i is given by

800 = 36a10 ,i + 940ν10 where ν =1

1 + iThe net income each year is £36 and this is 4.5% of the purchase price. There is a further gain on redemption. Hencethe net yield is greater than 4.5%. The gain on redemption is £140; if this amount was paid in equal instalmentsover the 10 years, then the payment each year would be £50, which is 6.25% of the purchase price (this is theapproximation in equation (1.3b). Hence the net yield i satisfies 4.5 < i < 6.25. Now use tables:

if i = 0.05, 36a10 ,i + 940ν10 = 847.339; if i = 0.055, 36a10 ,i + 940ν10 = 821.660; and if i = 0.06,36a10 ,i + 940ν10 = 789.854.

Interpolating between i = 0.055 and i = 0.06 gives:(x− 0.055)/(0.06− 0.055) = (f (x)− f (0.055))/(f (0.06)− f (0.055))

and hence x = 0.055 + 0.005(800− 821.660)/(789.854− 821.66) = 0.0584 or 5.8%.

1.6 Effect of term to redemption on the yield. Now consider how the yield from a bond, i, varies with n ifthe price of the bond is fixed. Equation (1.5c) has the following consequence (see exercise 2):

Suppose bonds A and B have the same redemption value C, the same annual coupon fr, the samepurchase price P and the same coupon r% per annum payable m times per year.Suppose bond A is redeemed after n1 years and bond B is redeemed after n2 years where n1 < n2. Leti1 and i2 denote the yields of bonds 1 and 2 respectively.• If P < C then i1 > i2 and so bond A with the shorter term has the higher yield.• If P > C then i1 < i2 and so bond B with the longer term has the higher yield.• If P = C then i1 = i2.

Informally, the result is obvious: if P < C, then the investor receives a capital gain of C − P when the bondis redeemed. It is clearly better to receive this amount sooner rather than later.

1.7 Optional redemption date. Sometimes a security does not have a fixed redemption date. The redemptiondate may be at the borrower’s option4. This could be at any date, after a specified date or at any one of a seriesof dates between two specified dates. The last possible redemption date is called the final redemption date; ifthere is no such date then the security is said to be undated. For a bond, the borrower is the issuer of the bond.

An investor who purchases a bond with a variable redemption rate at the option of the borrower (or issuer)cannot know the yield that will be obtained. It is possible to specify the following two amounts:

• The maximum price P to obtain a yield which is at least i0. The quantities f , r, t1, C andm are fixed knownnumbers. Suppose the bond can be redeemed after n years for any n satisfying n1 ≤ n ≤ n2 (where both n1and n2 are multiples of 1/m and n2 is possibly infinite).We think of P as a function such that if we are told n and i, then we can find P (n, i) by evaluating equa-tion (1.5c).

4 It is also possible for a security to be redeemable at the lender’s option, but this is much less common.


First case: suppose i(m) < (1− t1)g, equivalently, suppose P (n, i) > C.Then for a fixed price, a longer term implies a higher yield—see paragraph 1.6 above. So suppose n1 < n∗

and P (n1, i) = P (n∗, i∗) for some i and i∗; then i∗ > i.This means that whatever price we pay for the bond, the yield if the bond is redeemed at any date n∗ withn∗ > n1 must be greater than i, the yield if it is redeemed at the earliest time n1.

If i(m) < (1− t1)g, the purchaser of the bond will receive no capital gain and should price the bond onthe assumption that redemption will take place at the earliest possible time n1. Then, no matter whenthe bond is actually redeemed, the yield that will be obtained will be at least the value of the yield calculatedby assuming the bond is redeemed at the earliest possible time, n1.

Second case: suppose i(m) > (1− t1)g, equivalently suppose P (n, i) < C.Then for a fixed price, a shorter term implies a higher yield—see paragraph 1.6 above. So suppose n∗ < n2and P (n2, i) = P (n∗, i∗) for some i and i∗; then i∗ > i.

If i(m) > (1 − t1)g, the purchaser of the bond will receive a capital gain and should price the bond onthe assumption that redemption will take place at the latest possible time. Then, no matter when thebond is actually redeemed, the yield that will be obtained will be at least the value of the yield calculated byassuming the bond is redeemed at the latest possible time.

Third case: suppose i(m) = (1− t1)g, equivalently suppose P (n, i) = C for all n.If i(m) = (1− t1)g, the yield is same whatever the redemption date.• The minimum yield i if the price is P . Let P and C denote the price and redemption price per £100 facevalue, respectively. By the above remarks, we have the following:

If P < C then it is better for the investor that the bond is redeemed sooner rather than later. Suppose i2denotes the yield if the bond is redeemed at the latest possible time n2; then the yield will be at least i2whatever the redemption date.If P > C then it is better for the investor that the bond is redeemed later rather than sooner. Suppose i1denotes the yield if the bond is redeemed at the earliest possible time n1; then the yield will be at least i1whatever the redemption date.If P = C then the yield will be i where i(m) = (1− t1)g whatever the redemption date.

1.8 Clean and dirty prices. Suppose investor A sells a bond to another investor, B, between two coupondates. Then B will be the registered owner of the bond at the next coupon date and so receive the coupon. ButA will feel entitled to the “accrued coupon” or “accrued interest”—this is the interest A earned by holding thebond from the previous coupon date to the date of the sale. He therefore sells the bond for the dirty price—thisis the NPV of future cash flows from the bond5.The clean price is defined to be equal to the dirty price minus the accrued interest. So we have

dirty price = NPV of future cash flowsclean price = dirty price − accrued interest

The price quoted in the market is the clean price but the bond is sold for the dirty price.The way that accrued interest is calculated depends on the bond market. For U.S. treasury bonds, the conventionis ACT/ACT. Hence:

accrued interest = value of next coupon× days since last coupondays between coupons

For eurobonds, the convention is 30/360. This implies that the accrued interest on the 31st of the month isthe same at the accrued interest on the 30th of the month. Also, the accrued interest for the 7 days from 22February to 1 March would be 9/360 of the annual coupon. In the UK and Japan, the convention is ACT/365.Hence:

accrued interest = annual coupon× days since last coupon365

Example 1.8a. A bond has a semi-annual coupon with payment dates of 27 April and 27 October. Suppose thecoupon rate is 10% per annum convertible semi-annually. Calculate the accrued interest per £100 if the bond ispurchased on 4 July. Assume ACT/365.

5 Of course, this NPV depends on the current yield which in turn depends on the general market sentiment about the bond.


Solution. Number of days from 27 April to 4 July is 3 + 31 + 30 + 4 = 68. Hence accrued interest is 10× 68/365 =1.863.

Example 1.8b. A bond has a 9% coupon paid annually on August 11. Maturity is on 11 August 2005. The currentmarket yield for the bond is 8%. Find the dirty price and the clean price on (a) 8 June 2000; (b) 1 August 2000.Assume 30/360.

Solution. (a) Using 30/360 gives 63 days between 8/06/00 and 11/08/00.

Time 8/06/00 11/08/00 11/08/01 11/08/02 11/08/03 11/08/04 11/08/05Cash flow −P 9 9 9 9 9 109

The price on 11/08/00 is given by 9(1 + ν + ν2 + ν3 + ν4 + ν5) + 100ν5 where ν = 1/1.08. Hence the dirty price on8/06/00 (which is the NPV of future cash flows on 8/06/00) is given by

P = ν63/360[

9(

1− ν6

1− ν

)+ 100ν5

]= 111.48

The number of days from 11/08/99 to 8/06/00 is 297. Hence the accrued interest is 9 × 297/360 = 7.425. Hencethe clean price is 111.48− 7.42 = 104.06. The clean price is the price quoted in the markets and financial press.For part (b), we have 10 days instead of 63 and 350 days instead of 297. This gives a dirty price of 112.75, accruedinterest of 8.75 and a clean price of 104.00.

If a bond is quoted as cum dividend then the buyer will receive the next coupon payment; if a bond is quotedas ex dividend then the buyer will not receive the next coupon payment (and the accrued interest is thennegative). UK gilts normally go ex dividend 7 working days before a coupon payment.

.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

........ ............ ............

time

................................................................

................................................................

................................................................

.....................................................................................

......................................................................

......................................................................

......................................................................

......................................................................

......................................................................

.........................................

............................................................... ............

..........................

cleanprice

cum dividend ex dividend

coupon coupondate date

Figure 1.8a. Plot of dirty price against time(PICTEX)

1.9

Summary. Bond calculations.• Suppose coupon rate is nominal r% p.a. payable m times per year and i is the effective yield perannum. Let g = annual coupon/redemption value = fr/C. Then the price P is:

P = fr(1− t1)a(m)n,i +

C

(1 + i)n=g(1− t1)i(m)

[C − Cνn

]+ Cνn

• Gross running yield =annual couponcurrent price

=fr

P;

• Net running yield =annual coupon after tax

current price=

(1− t1)frP

• Gross redemption yield (or yield to maturity) and net redemption yield.• Effect of capital gains tax on yield. Suppose i is the yield.

If i(m) > (1− t1)g then capital gains tax is payable.• Optional redemption date: at option of issuer.

If i(m) < (1− t1)g there is no capital gain; purchaser should assume early redemption.If i(m) > (1− t1)g there is a capital gain; purchaser should assume late redemption.

• Clean and dirty price.

5 Bonds, Equities and Inflation Sep 27, 2016(9:51) Exercises 2 Page 87


1. Consider a bond with face value f , redemption value C, current price P which has a coupon rate of r% payablem times per year for n years. Taxation is at the rate t1 and is paid at the end of each year—hence the annual tax ist1fr. Find an equation for the yield to maturity.

2. Suppose bond A is redeemed after n1 years and bond B is redeemed after n2 years where n1 < n2. Suppose furtherthat both bonds have the redemption value C, the same annual coupon fr, the same purchase price P and both havea coupon which is payable m times per year.Let i1 and i2 denote the yields of bonds 1 and 2 respectively. Show the following:(a) If P = C then i1 = i2. (b) If P < C then i1 > i2. (c) If P > C then i1 < i2.

3. Suppose the first coupon on an undated bond with current price P and face value f is delayed for t0 years. Thecoupon rate is r% nominal p.a. payable m times per year and the income tax rate is t1.Show that the effective annual yield, i, is given by

P = fr(1− t1)νt0 a(m)∞ =

fr(1− t1)νt0

d(m) where ν =1

1 + i

4. A bond with an annual coupon of 8% per annum has just been issued with a gross redemption yield of 6% per annumeffective. It is redeemable at par at the option of the borrower on any coupon payment date from the tenth anniversaryof issue to the twentieth anniversary of issue. The gross redemption yield on bonds of all terms to maturity is 6% perannum effective.Ten years after issue, the gross redemption yield on bonds of all terms to maturity is 10% per annum effective.(a) Is the bond likely to be redeemed earlier or later than was assumed at issue?(b) Is the gross redemption yield likely to be higher or lower than was assumed at issue?

(Institute/Faculty of Actuaries Examinations, September 1998 (adapted)) [3]

5. An investor purchases a bond, redeemable at par, which pays half-yearly coupons at a rate of 8% per annum. Thereare 8 days until the next coupon payment and the bond is ex-dividend. The bond has 7 years to maturity after thenext coupon payment.Calculate the purchase price to provide a yield to maturity of 6% per annum.

(Institute/Faculty of Actuaries Examinations, April, 2002) [4]

6. A fixed interest stock bears a coupon of 7% per annum payable half yearly on 1 April and 1 October. It is redeemableat par on any 1 April between 1 April 2004 and 1 April 2010 inclusive at the option of the borrower.On 1 July 1991 an investor purchased £10,000 nominal of the stock at a price to give a net yield of 6% per annumeffective after allowing for tax at 25% on the coupon payments.On 1 April 1999 the investor sold the holding at a price which gave a net yield of 5% per annum effective to anotherpurchaser who is also taxed at a rate of 25% on the coupon payments.

(i) Calculate the price at which the stock was bought by the original investor.(ii) Calculate the price at which the stock was sold by the original investor.


7. An investor purchases a bond on the issue date at a price of £96 per £100 nominal. Coupons at an annual rate of 4%are paid annually in arrears. The bond will be redeemed at par 20 years after the issue date.Calculate the gross redemption yield from the bond.


8. A new issue of a fixed interest security has a term to redemption of 20 years and is redeemable at 110%. The securitypays a coupon of 91/2% per annum payable half-yearly in arrears.An investor who is liable to tax on all income at a rate of 23% and on all capital gains at a rate of 34% bought allthe stock at the date of issue at a price which gave the investor a yield to maturity of 8% per annum effective. Whatprice did the investor pay per £100 nominal of the stock?


9. (i) Describe the risk characteristics of a government-issued, conventional, fixed-interest bond.(ii) A particular government bond is structured as follows.

Annual coupons are paid in arrears of 8% of the nominal value of the bond. After 5 years, a capitalrepayment is made, equal to half of the nominal value of the bond. The capital is repaid at par. Therepayment takes place immediately after the payment of the coupon due at the end of the 5th year. Afterthe end of the 5th year, coupons are only paid on that part of the capital that has not been repaid. At the endof the 10th year, all the remaining capital is repaid.

Calculate the purchase price of the bond per £100 nominal, at issue, to provide a purchaser with an effective netrate of return of 6% per annum. The purchaser pays tax at a rate of 30% on coupon payments only.



10. A loan of nominal amount of £100,000 is to be issued bearing coupons payable quarterly in arrear at a rate of 5%per annum. Capital is to be redeemed at 103 on a single coupon date between 15 and 20 years after the date of issue,inclusive. The date of redemption is at the option of the borrower.An investor who is liable to income tax at 20% and capital gains tax of 25% wishes to purchase the entire loan atthe date of issue. Calculate the price which the investor should pay to ensure a net effective yield of at least 4% perannum. (Institute/Faculty of Actuaries Examinations, April 2005) [9]

11. A loan of nominal amount £100,000 is to be issued bearing interest payable half-yearly in arrears at a rate of 7% perannum. The loan is to be redeemed with a capital payment of 110 per £100 nominal on a coupon date between 10and 15 years after the date of issue, inclusive, the date of redemption being at the option of the borrower.An investor who is liable to income tax at 25% and capital gains tax of 30% wishes to purchase the entire loan at thedate of issue at a price which ensures that the investor achieves a net effective yield of at least 5% per annum.

(i) Determine whether the investor would make a capital gain if the investment is held until redemption.(ii) Explain how your answer to (i) influences the assumptions made in calculating the price the investor should pay.

(iii) Calculate the maximum price which the investor should pay.(Institute/Faculty of Actuaries Examinations, September 2002) [3+2+5=10]

12. An investor purchased a bond with exactly 15 years to redemption. The bond, redeemable at par, has a grossredemption yield of 5% per annum effective. It pays coupons of 4% per annum, half yearly in arrear. The investorpays tax at 25% on the coupons only.

(i) Calculate the price paid for the bond.(ii) After exactly 8 years, immediately after the payment of the coupon then due, this investor sells the bond to

another investor who pays income tax at a rate of 25% and capital gains tax at a rate of 40%. The bond ispurchased by the second investor to provide a net return of 6% per annum effective.(a) Calculate the price paid by the second investor.(b) Calculate, to one decimal place, the annual effective rate of return earned by the first investor during the

period for which the bond was held.(Institute/Faculty of Actuaries Examinations, September 2005) [3+10=13]

13. Ten million pounds nominal of loan stock was issued on 1 March 1990. The stock is due to be redeemed in onepayment of 110% nominal on 1 March 2000. The stock pays interest at 10% per annum, payable half-yearly inarrears on 1 September and 1 March.

(i) An investor, subject to income tax at 25% but not liable to capital gains tax, bought £10,000 nominal of thestock at issue so as to obtain a net effective yield of 8% per annum, after tax. What price did the investor payper £100 nominal on the date of issue?

(ii) On 1 March 1998 the first investor, having received the interest due on that date, sold the stock to a secondinvestor who was subject to 40% tax on both income and capital gains. The price paid by the second investorwas calculated so as to earn him a net effective yield of 6% per annum, after tax. What price did the secondinvestor pay per £100 nominal?

(iii) Assuming that the first investor sold the whole of the stock of his loan to the second investor, what was theeffective net yield, after tax, earned on the first investment?


14. An investor purchased a bond with exactly 20 years to redemption. The bond, redeemable at par, has a grossredemption yield of 6%. It pays annual coupons, in arrears, of 5%. The investor does not pay tax.

(i) Calculate the purchase price of the bond.(ii) After exactly 10 years, immediately after payment of the coupon then due, this investor sells the bond to another

investor. That investor pays income and capital gains tax at a rate of 30%. The bond is purchased by the secondinvestor to provide a net rate of return of 6.5% per annum.(a) Calculate the price paid by the second investor.(b) Calculate the annual effective rate of return earned by the first investor during the period for which the

bond was held. (Institute/Faculty of Actuaries Examinations, September, 2001) [3+10=13]

15. A fixed interest security pays coupons of 8% per annum half yearly on 1 January and 1 July. The security willbe redeemed at par on any 1 January between 1 January 2006 and 1 January 2011 inclusive, at the option of theborrower.An investor purchased a holding of the security on 1 January 2001, immediately after the payment of the couponthen due, at a price which gave him a net yield of at least 5% per annum effective. The investor pays tax at 40% oninterest income and 30% on capital gains. On 1 January 2003 the investor sold the holding, immediately after thepayment of the coupon then due, to a fund which pays no tax at a price to give the fund a gross yield of at least 7%per annum effective.

(i) Calculate the price per £100 nominal at which the investor bought the security.(ii) Calculate the price per £100 nominal at which the investor sold the security.

(iii) Calculate the net yield per annum convertible half-yearly, which the investor actually received over the twoyears the investor held the security. (Institute/Faculty of Actuaries Examinations, April 2003) [5+3+6=14]


16. A loan of nominal amount £10,000,000 is to be issued bearing a coupon of 8% per annum payable quarterly inarrears. The loan is to be repaid at the end of 15 years at 110% of the nominal value. An institution not subject toeither income or capital gains tax bought the whole issue at a price to yield 9% per annum effective.

(i) Calculate the price per £100 nominal which the institution paid.(ii) Exactly 5 years later, immediately after the coupon payment, the institution sold the entire issue of stock to an

investor who pays both income and capital gains tax at a rate of 20%. Calculate the price per £100 nominalwhich the investor pays the institution to earn a net redemption yield of 7% per annum effective.

(iii) Calculate the net running yield per annum earned by this investor.(iv) Calculate the annual effective rate of return earned by the institution in (i).


17. A loan is issued bearing interest at a rate of 9% per annum and payable half-yearly in arrear. The loan is to beredeemed at £110 per £100 nominal in 13 years’ time.

(i) The loan is issued at a price such that an investor, subject to income tax at 25% and capital gains tax at 30%,would obtain a net redemption yield of 6% per annum effective. Calculate the issue price per £100 nominal ofthe stock.

(ii) Two years after the date of issue, immediately after a coupon payment has been made, the investor decides tosell the stock and finds a potential buyer who is subject to income tax at 10% and capital gains tax at 35%. Thepotential buyer is prepared to buy the stock provided she will obtain a net redemption yield of at least 8% perannum effective.(a) Calculate the maximum price (per £100 nominal) which is the original investor can expect to obtain from

the potential buyer.(b) Calculate the net effective annual redemption yield (to the nearest 1% per annum effective) that will be

obtained by the original investor if the loan is sold to the buyer at the price determined in (ii)(a).(Institute/Faculty of Actuaries Examinations, April 2007) [5+10=15]

18. A loan of nominal amount of £100,000 is to be issued bearing interest payable quarterly in arrear at a rate of 8%. p.a.Capital is to be redeemed at £105% on a coupon date between 15 and 20 years after the date of issue, inclusive, thedate of redemption being at the option of the borrower.

(i) An investor who is liable to income tax at 40% and tax on capital gains at 30% wishes to purchase the entireloan at the date of issue. What price should she pay to ensure a net effective yield of at least 6% p.a?

Exactly 10 months after issue the loan is sold to an investor who pays income tax at 20% and capital gains at30%. Calculate the price this investor should pay to achieve a yield of 6% p.a. on the loan:(a) assuming redemption at the earliest possible date(b) assuming redemption at the latest possible date.

(ii) Explain which price this second investor should pay to achieve a yield of at least 6% p.a.(Institute/Faculty of Actuaries Examinations, April 1997) [17]

19. A fixed interest security pays coupons of 8% per annum half yearly on 1 January and 1 July. The security will beredeemed at par on any 1 January from 1 January 2017 to 1 January 2022 inclusive, at the option of the borrower.An investor purchases a holding of the security on 1 May 2011, at a price which gave him a net yield of at least6% per annum effective. The investor pays tax at 30% on interest income and 25% on capital gains.On 1 April 2013 the investor sold the holding to a fund which pays no tax at a price to give the fund a gross yield ofat least 7% per annum effective.

(i) Calculate the price per £100 nominal at which the investor bought the security.(ii) Calculate the price per £100 nominal at which the investor sold the security.

(iii) Show that the effective net yield that the investor obtained on the investment was between 8% and 9% perannum. (Institute/Faculty of Actuaries Examinations, April 2013) [5+3+6=14]

20. A loan of nominal amount £100,000 is to be issued bearing coupons payable quarterly in arrear at a rate of 7%per annum. Capital is to be redeemed at 108% on a coupon date between 15 and 20 years after the date of issue,inclusive. The date of redemption is at the option of the borrower.An investor who is liable to income tax at 25% and capital gains tax at 35% wishes to purchase the entire loan at thedate of issue.Calculate the price which the investor should pay to ensure a net effective yield of at least 5% per annum.


21. A fixed-interest bond pays annual coupons of 5% per annum in arrear on 1 March each year and is redeemed at paron 1 March 2025. On 1 March 2007, immediately after the payment of the coupon then due, the gross redemptionyield was 3.158% per annum effective.

(i) Calculate the price of the bond per £100 nominal on 1 March 2007.On 1 March 2012, immediately after the payment of the coupon then due, the gross redemption yield on the bondwas 5% per annum.(ii) State the new price of the bond per £100 nominal on 1 March 2012.


A tax-free investor purchased the bond on 1 March 2007, immediately after payment of the coupon then due, andsold the bond on 1 March 2012 immediately after payment of the coupon then due.(iii) Calculate the gross annual rate of return achieved by the investor over this period.(iv) Explain, without doing any further calculations, how your answer to part (iii) would change if the bond were

due to be redeemed on 1 March 2035 (rather than 1 March 2025). You may assume that the gross redemptionyield at both the date of purchase and the date of sale remains the same as in parts (i) and (ii) above.


22. A fixed interest security pays coupons of 4% per annum, half-yearly in arrear and will be redeemed at par in exactlyten years.

(i) Calculate the price per £100 nominal to provide a gross redemption yield of 3% per annum convertible halfyearly.

(ii) Calculate the price, 91 days later, to provide a net redemption yield of 3% per annum convertible half-yearly, ifincome tax is payable at 25%. (Institute/Faculty of Actuaries Examinations, September 2013) [2+2=4]

23. An investor is considering the purchase of two government bonds, issued by two countries A and B respectively,both denominated in euro.Both bonds provide a capital repayment ofe100 together with a final coupon payment ofe6 in exactly one year. Theinvestor believes that he will receive both payments from the bond issued by country A with certainty. He believesthat there are four possible outcomes for the bond from country B, shown in the table below.

Outcome ProbabilityNo coupon or capital payment 0.1Capital payment received, but no coupon payment received 0.250% of capital payment received, but no coupon payment received 0.3Both coupon and capital payments received in full 0.4

The price of the bond issued by country A is e101.(i) Calculate the price of the bond issued by country B to give the same expected return as that for the bond issued

by country A.(ii) Calculate the gross redemption yield from the bond issued by country B assuming that the price is as calculated

in part (i).(iii) Explain why the investor might require a higher expected return from the bond issued by country B than from

the bond issued by country A. (Institute/Faculty of Actuaries Examinations, September 2013) [3+1+2=6]

3 Equity calculations3.1 Calculating the yield. Owners of equities receive a stream of dividend payments. The sizes of thesedividend payments are uncertain and depend on the performance of the company. In the UK they are oftenpaid six-monthly with an interim and a final dividend.

Consider the following special case of annual dividends (the results for six-monthly dividend payments areobtained in the same manner). Suppose the present price of a share just after the annual dividend has been paidis P . Suppose dividends are paid annually and dk is the estimated gross dividend in k years time. Then thecash flow is displayed in the following table:

Time 0 1 2 3 . . .Cash flow −P d1 d2 d3 . . .

The yield i of the share given by:

P =∞∑k=1

dk(1 + i)k

If dividends are assumed to grow by a constant proportion each year, then the cash flow is:Time 0 1 2 3 . . .Cash flow −P d1 d1(1 + g) d1(1 + g)2 . . .

and so the yield i of the share given by:

P =∞∑k=1

d1(1 + g)k−1

(1 + i)k=

d1

i− g


Example 3.1a. The next dividend on an equity is due in 6 months and is expected to be 6 pence. Subsequentdividends will be paid annually. Assume the second, third and fourth dividends are each 10% greater than theirpredecessor. Assume also that dividends subsequent to the fourth grow at the rate of 5% per annum. Calculate theNPV of the dividend stream assuming the investor’s yield is 12% per annum.

Solution. The cash flow is as follows:Time 0 1/2 3/2 5/2 7/2 9/2 . . .Cash flow (in pence) 0 6 6× 1.1 6× 1.12 6× 1.13 6× 1.13 × 1.05 . . .

Let ν = 1/(1 + i) = 1/1.12. Hence

NPV = 6ν1/2 + 6(1.1)ν3/2 + 6(1.1)2ν5/2 + 6(1.1)3ν7/2 +∞∑k=5

6(1.1)3(1.05)k−4ν(2k−1)/2

= 6ν1/2 + 6(1.1)ν3/2 + 6(1.1)2ν5/2 + 6(1.1)3ν7/2 + 6(1.1)3ν9/2 1.051− 1.05ν

3.2 Other measures. Estimating future dividends involves a large amount of subjectivity. Consequently,other numerical measures for comparing equities are usually used—mainly the dividend yield and the P/Eratio.The dividend yield (often described simply as the yield) is given by

Dividend yield =gross annual dividend per share

current share priceThe dividend yield is an indication of the current level of income from a share.The P/E ratio or price/earnings ratio is given by

P/E ratio =current share price

net annual profit per shareThe net annual profit per share is the last annual profits of the company divided by the number of shares. Forexample, suppose the share price is 400p and the earnings per share is 20p. Then the P/E ratio is 20; i.e. theshares are on a multiple of 20, or the shares sell at 20 times earnings.

Yields and P/E ratios move in opposite directions. If the share price rises, the yield falls and the P/E ratio rises.Typically a high P/E ratio suggests a growth company and a low P/E ratio suggests a company with more staticprofits. (Belief that a company is a growth company implies belief in higher dividends in the future and hencea higher share price.) Further information about these measures is given in other courses.

4 Real and money rates of interest

4.1 Idea. The real rate of interest is the rate of interest after allowing for inflation; the money rate of interestis the rate of interest without allowing for inflation.

Example 4.1a. An amount of £100 grows to £120 in 1 year. Inflation is 5% p.a. Find the money and real rates ofinterest.

Solution. The money rate of interest is 20% p.a. In real terms, £100 grows to £120/1.05 = £114.29 in 1 year. Hencethe real rate of interest is 14.29% p.a.

Suppose iM denote the money rate of interest, iR denotes the real rate of interest and q denotes the inflationrate, then

1 + iM = (1 + q)(1 + iR) or, equivalently, iR =iM − q1 + q

(4.1a)

Note that iM = iR(1 + q) + q ≈ iR + q if iR and q are small.

Suppose £1 grows to £A(t) over the interval (0, t). The amount £A(t) may be worth only £A∗(t) at time 0 afterallowing for inflation, where A∗(t) < A(t). The rate of interest needed for £1 to grow to £A∗(t) is the real rateof interest and is, in this case, less than the money rate of interest.The term deflation means that the rate of inflation is negative and A∗(t) > A(t). The real rate of interest isthen higher than the money rate of interest.


Example 4.1b. Find the real rate of interest for the following cash flow using the inflation index Q(t):Time 0 1 2 3Cash flow a0 = −100 a1 = 8 a2 = 8 a3 = 108Inflation Index, Q(t) 150 156 166 175

Solution. The cash flow should be adjusted as follows:

Cash flow a0 = −100 a1 = 8× Q(0)Q(1) a2 = 8× Q(0)

Q(2) a3 = 108× Q(0)Q(3)

This leads to a real rate of interest iR given by

100 =8× 150

156νR +

8× 150166

ν2R +

108× 150175

ν3R where νR =

11 + iR

and hence iR = 2.63

In general, consider the cash flowTime 0 t1 t2 . . . tnCash flow 0 Ct1 Ct2 . . . CtnInflation Index Q(0) Q(t1) Q(t2) . . . Q(tn)

The equation for the yield is given byn∑k=1

CtkQ(0)Q(tk)

νtk = 0 which leads ton∑k=1

CtkQ(tk)

νtk = 0 (4.1b)

where, as usual, ν = 1/(1 + i) and i is the yield. Equation (4.1b) shows that the yield does not depend on thebase value of the inflation index.

4.2 Special case: constant inflation assumption. Suppose we assume that the rate of inflation will be q%per annum. Consider the following cash flow:

Time 0 t1 t2 . . . tnCash flow 0 Ct1 Ct2 . . . CtnInflation Index 1 (1 + q)t1 (1 + q)t2 . . . (1 + q)tn

The real yield iR is given byn∑k=1

Ctk(1 + q)tk

νtkR = 0 where νR =1

1 + iR

The money yield iM is given byn∑k=1

CtkνtkM = 0 where νM =

11 + iM

Note that νR = (1 + q)νM which is just another version of equation (4.1a): (1 + iR)(1 + q) = 1 + iM .

4.3 Inflation linked payments. Suppose we have the series of payments c = ct1 , . . . , ctn and the paymentctk due at time tk is adjusted in line with some inflation index Q(t). This means that the payment at time tkwill become

cq,tk = ctkQ(tk)Q(0)

Hence the NPV of the series of inflation adjusted payments cq = cq,1, . . . , cq,n using the rate of interest i is

NPVi(cq) =n∑k=1

cq,tkνtk where ν =

11 + i

Consider adjusting the interest rate for inflation. If iR denotes the real rate of interest corresponding to themoney rate of interest iM , then 1 + iM = (1 + iR)Q(1)/Q(0) and in general (1 + iM )tk = (1 + iR)tkQ(tk)/Q(0).This implies that even if iM is constant over time, iR is a function of t.

This result generalises as follows: suppose we have both a discrete payment stream, c, and a continuouspayment stream, ρ. If NPVi(c, ρ) denotes the net present value at time 0 using interest rate i, then we knowthat

NPVi(c, ρ) =∑k

ctkνtk +

∫ ∞0

ρ(u)νu du


Now suppose every cash payment is adjusted to allow for inflation at the rate of q per unit time, then

NPVi(cq, ρq) =∑k

ctk (1 + q)tkνtk +∫ ∞

0ρ(u)(1 + q)uνu du

Using the relation 1 + iM = (1 + q)(1 + iR), or νR = (1 + q)νM , shows thatNPViM (cq, ρq) = NPViR(c, ρ) where (1 + iM ) = (1 + iR)(1 + q). (4.3a)

The NPV using the real rate of interest and payments not adjusted for inflation is the same as the NPV ofinflation adjusted payments using the money rate of interest.If i0 is the yield of a project when inflation is ignored, then NPVi0(c, ρ) = 0. Using equation (4.3a) implies thatNPVi1(cq, ρq) = 0 where (1 + i1) = (1 + i0)(1 + q). Hence the yield of the project when payments are adjustedfor inflation is i1 = (1 + q)i0 + q.Projects which appear unprofitable when interest rates are high, may become profitable when only a smallallowance is made for inflation.

4.4 Index linked bonds. Clearly, if the payments on a bond are linked precisely to the inflation index, thenthe real yield of the bond will be the same whatever happens to the the rate of inflation. However, sometimesthe calculation is more complicated than this. Index linked UK government securities have coupons linked tothe value of the inflation index 8 months before the payment is made. The real yield must be calculated usingthe inflation index at the times the actual payments are made.

Example 4.4a. Suppose an index-linked bond with face value f is redeemable after n years with redemptionvalue C. The coupon rate is r% per annum convertible semi-annually. The coupon payments are adjusted by usingthe inflation index Q. If the current price of the bond is P , find an equation for the money yield iM .

Solution. The cash flow is as follows:Time 0 1/2 1 3/2 . . . (2n− 1)/2 nCash flow (unadjusted) −P fr/2 fr/2 fr/2 . . . fr/2 fr/2 + CInflation Index Q(0) Q(1/2) Q(1) Q(3/2) . . . Q(n− 1/2) Q(n)

The equation for the money yield, iM is

P =2n∑k=1

fr

2Q(k/2)Q(0)

νk/2M + C

Q(n)Q(0)

νnM

4.5

Summary.• Equities: calculating the yield from forecasted dividends; the dividend yield; the P/E ratio.• Real rate of interest and money rate of interest: (1 + iM ) = (1 + q)(1 + iR).• Adjusting coupon payments using an inflation index.


1. On the assumption of constant future price inflation of 2.5% per annum, an investor has purchased a UK governmentindex-linked bond at a price to provide a real rate of return of 2% per annum effective if held to redemption. Explainwhy the real yield to redemption will be lower if the actual constant rate of inflation is higher than the assumed 2.5%per annum. (Institute/Faculty of Actuaries Examinations, April, 2002) [2]

2. An investor has earned a money rate of return from a portfolio of bonds in a particular country of 1% per annumeffective over a period of 10 years. The country has experienced deflation (negative inflation) of 2% per annumeffective during the period. Calculate the real rate of return per annum over the 10 years.


3. Dividends payable on a certain share are assumed to increase at a compound rate of 3% per half-year. A dividendof £5 per share has just been paid. Dividends are paid half-yearly. Find the value of the share to the nearest £1,assuming an effective rate of interest of 8% p.a. (Institute/Faculty of Actuaries Examinations, May 1999) [3]

4. Dividends payable on a share are assumed to increase at a compound rate of 21/2% per half-year. A dividend of £2per share has just been paid. Dividends are payable half-yearly. Find the value of the share to the nearest £1 assumingan effective rate of interest of 7% per annum. (Institute/Faculty of Actuaries Examinations, April 1999) [3]


5. (i) Define the characteristics of a government index-linked bond.(ii) Explain why most index-linked securities issued carry some inflation risk in practice.


6. An investor has invested a sum of £10m. Exactly one year later, the investment is worth £11.1m. An index of priceshas a value of 112 at the beginning of the investment and 120 at the end of the investment. The investor pays tax at40% on all money returns from investment. Calculate:(a) The money rate of return per annum before tax.(b) The rate of inflation.(c) The real rate of return per annum after tax. (Institute/Faculty of Actuaries Examinations, September 2006) [4]

7. An investor purchased a holding of ordinary shares two months before payment of the next dividend was due.Dividends are paid annually and it is expected that the next dividend will be a net amount of 12p per share. Theinvestor anticipates that dividends will grow at a constant rate of 4% per annum in perpetuity.Calculate the price per share that the investor should pay to obtain a net return of 7% per annum effective.


8. An investor is considering the purchase of 100 ordinary shares in a company. Dividends from the share will be paidannually. The next dividend is due in one year and is expected to be 8p per share. The second dividend is expectedto be 8% greater than the first dividend and the third dividend is expected to be 7% greater than the second dividend.Thereafter dividends are expected to grow at 5% per annum compound in perpetuity.Calculate the present value of this dividend stream at a rate of interest of 7% per annum effective.


9. A loan of £20,000 was issued, and was repaid at par after 3 years. Interest was paid on the loan at the rate of 10%per annum, payable annually in arrears. The value of the retail price index at various times was as follows:

At the date the loan was made 245.0One year later 268.2Two years later 282.2Three years later 305.5

Calculate the real rate of return earned on the loan. (Institute/Faculty of Actuaries Examinations, April 1997) [5]

10. An equity pays annual dividends and the next dividend, payable in 10 months time, is expected to be 5p. Thereafter,dividends are expected to grow by 3% per annum compound. Inflation is expected to be 2% per annum throughout.Calculate the value of the equity assuming that the real yield obtained is 2.5% per annum, convertible half-yearly.


11. An investment which pays annual dividends is bought immediately after a dividend payment has been made. Divi-dends are expected to grow at a compound rate of g per annum and price inflation is expected to be at a rate of e perannum. If the dividend payment expected at the end of the first year is d per unit invested and if it is assumed theinvestment is held indefinitely, show that the expected real rate of return per annum per unit invested is given byi′ = (d + g − e)/(1 + e). (Institute/Faculty of Actuaries Examinations, September 1999) [5]

12. An ordinary share pays annual dividends. The next dividend is expected to be 10p per share and is due in exactly9 months’ time. It is expected that subsequent dividends will grow at a rate of 5% per annum compound and thatinflation will be 3% per annum. The price of the share is 250p and dividends are expected to continue in perpetuity.Calculate the expected effective real rate of return per annum for an investor who purchases the shares.


13. A share has a price of 750p and has just paid a dividend of 30p. The share pays annual dividends. Two analysts agreethat the next expected dividend will be 35p but differ in their estimates of long-term dividend growth. One analystexpects no dividend growth and the other analyst expects constant dividend growth of 10% per annum, after the nextdividend payment.

(i) Derive a formula for the value of a share in terms of the next expected dividend (d1), a constant rate of dividendgrowth (g), and the rate of return from the share (i).

(ii) Calculate the rates of return the two analysts expected from the share.(Institute/Faculty of Actuaries Examinations, September 1999) [3+3=6]

14. An ordinary share pays annual dividends. The next dividend is expected to be 5p per share and is due in exactly3 months’ time. It is expected that subsequent dividends will grow at a rate of 4% per annum compound and thatinflation will be 1.5% per annum. The price of the share is 125p and dividends are expected to continue in perpetuity.Calculate the effective real rate of return per annum for an investor who purchases the share.



15. An investor is considering investing in the shares of a particular company.The shares pay dividends every 6 months, with the next dividend being due in exactly 4 months’ time. The nextdividend is expected to be d1, the purchase price of a single share is P , and the annual effective rate of returnexpected from the investment is 100i%. Dividends are expected to grow at a rate of 100g% per annum from the levelof d1 where g < i. Dividends are expected to be paid in perpetuity.

(i) Show that

P =d1(1 + i)1/6

(1 + i)1/2 − (1 + g)1/2

(ii) The investor delays purchasing the shares for exactly 2 months at which time the price of a single share is £18,100g% = 4% and d1 = £0.50. Calculate the annual effective rate of return expected by the investor to the nearest1%. (Institute/Faculty of Actuaries Examinations, April 2001) [7]

16. (i) Describe the characteristics of ordinary shares (equities).(ii) The prospective dividend yield from an ordinary share is defined as the next expected dividend divided by the

current price. A particular share is expected to provide a real rate of return of 5% per annum effective. Inflationis expected to be 2% per annum. The growth rate of dividends from the share is expected to be 3% per annumcompound. Dividends will be paid annually and the next dividend is expected to be paid in 6 months’ time.Calculate the prospective dividend yield from the share.


17. The following investments were made on 15 January 1993.Investment A: £10,000 was placed in a special savings account with a 5-year term. Invested money was accumulatedat 31/2% per annum effective for the first year, 41/2% per annum effective for the second year, 51/2% per annumeffective for the third year, 61/2% per annum effective for the fourth year, and 71/2% per annum effective for the fifthyear.Investment B: £10,000 was placed in a zero coupon bond with a 5-year term in which the redemption proceeds werethe amount invested multiplied by the ratio of the Retail Price Index for the month two months prior to that in whichredemption fell to the Retail Price Index for the month two months prior to the date of investment; this amount wasfurther increased by 23/4% compound for each complete year money was invested.Investment C: An annuity payable annually in arrears for 5 years was purchased for £10,000 to yield 61/4% perannum effective.The Retail Price Index at various times was as follows:

November 1992 237.6January 1993 240.0January 1994 250.0January 1995 264.4January 1996 266.6January 1997 270.4November 1997 274.0January 1998 275.6

(i) What is the amount of the annual income yielded by the annuity?(ii) Calculate the real rate of return per annum earned on investment A and on investment B over the period 15 Jan-

uary 1993 to 15 January 1998. The investor is not liable for tax.(iii) Determine which of the three investments yielded the highest real rate of return per annum over the period

15 January 1993 to 15 January 1998. (Institute/Faculty of Actuaries Examinations, April 1998) [1+4+3=8]

18. An investor purchases a bond 3 months after issue. The bond will be redeemed at par ten years after issue and payscoupons of 6% per annum annually in arrears. The investor pays tax of 25% on both income and capital gains (withno relief for indexation).

(i) Calculate the purchase price of the bond per £100 nominal to provide the investor with a rate of return of 8% perannum effective.

(ii) The real rate of return expected by the investor from the bond is 3% per annum effective. Calculate the annualrate of inflation expected by the investor. (Institute/Faculty of Actuaries Examinations, April, 2002) [6+2=8]

19. In a particular country, income tax and capital gains tax are both collected on 1 April each year in relation to grosspayments made during the previous 12 months.A fixed interest bond is issued on 1 January 2003 with term of 25 years and is redeemable at 110%. The securitypays a coupon of 8% per annum, payable half-yearly in arrears.An investor, who is liable to tax on income at a rate of 25% and on capital gains at a rate of 30%, bought £10,000nominal of the stock at issue for £9,900.

(i) Assuming an inflation rate of 3% per annum over the term of the bond, calculate the net real yield obtained bythe investor if they hold the stock to redemption.

(ii) Without doing any further calculation, explain how and why your answer to (i) would alter if tax were collectedon 1 June instead of 1 April each year. (Institute/Faculty of Actuaries Examinations, April 2004) [9+2=11]


20. An ordinary share pays annual dividends. The next dividend is due in exactly 8 months’ time. This dividend isexpected to be £1.10 per share. Dividends are expected to grow at a rate of 5% per annum compound from this leveland are expected to continue in perpetuity. Inflation is expected to be 3% per annum. The price of the share is £21.50.Calculate the expected effective annual real rate of return for an investor who purchases the share.


21. A government issued a number of index-linked bonds on 1 June 2000 which were redeemed on 1 June 2002. Eachbond had a nominal coupon rate of 3% per annum, payable half-yearly in arrears, and a nominal redemption priceof 100. The actual coupon and redemption payments were indexed according to the increase in the retail price indexbetween 6 months before the bond issue date and 6 months before the coupon or redemption dates.The values of the retail price index in the relevant months were:

Date Retail Price IndexDecember 1999 100June 2000 102December 2000 107June 2001 111December 2001 113June 2002 118

(i) An investor purchased £100,000 nominal at the issue date, and held it until it was redeemed. The issue pricewas £94 per £100 nominal.Calculate all the investor’s cash flows from this investment, before tax.

(ii) The investor is subject to income tax at a rate of 25% and capital gains tax at a rate of 35%. When calculatingthe amount of capital gain which is subject to tax, the price paid for the investment is indexed in line with theincrease in the retail price index between the month in which the investment was purchased and the month inwhich it was redeemed.(a) Calculate the investor’s capital gains tax liability in respect of this investment.(b) Calculate the net effective yield per annum obtained by the investor on these bonds.


22. At time t = 0 an investor purchased an annuity-certain which paid her £10,000 per annum annually in arrear for threeyears. The purchase price paid by the investor was £25,000.The value of the retail price index at various times was as shown in the table below:

Time t (years) t = 0 t = 1 t = 2 t = 3Retail price index 170.7 183.3 191.0 200.9

(i) Calculate, to the nearest 0.1%, the following effective rates of return per annum achieved by the investor fromher investment in the annuity:(a) the real rate of return; (b) the money rate of return.

(ii) By considering the average rate of inflation over the three-year period, explain the relationship between youranswers in (a) and (b) in (i). (Institute/Faculty of Actuaries Examinations, April 2005) [7+2=9]

23. An ordinary share pays annual dividends. A dividend of 25p per share has just been paid. Dividends are expected togrow by 2% next year and by 4% the following year. Thereafter, dividends are expected to grow at 6% per annumcompound in perpetuity.

(i) State the main characteristics of ordinary shares.(ii) Calculate the present value of the dividend stream described above at a rate of interest of 9% per annum effective

from a holding of 100 ordinary shares.(iii) An investor buys 100 shares in (ii) for £8.20 each. He holds them for two years and receives the dividends

payable. He then sells them for £9 immediately after the second dividend is paid.Calculate the investor’s real rate of return if the inflation index increases by 3% during the first year and by 3.5%during the second year assuming dividends grow as expected.


24. On 9 October 1997 an investor, not liable to tax, had the choice of purchasing either:(A) 71/2% Treasury Stock at a price of £107 per £100 nominal repayable at par in 2006.(B) 2% Index Linked Treasury Stock 2006 at a price of £203 per £100 nominal. The RPI base figure for indexing

was 69.5 and the index applicable to the next coupon (payable on 8 April 1998) was 158.5. (This may also betaken as the latest known value of the RPI on 9 October.)

Assume both stocks are redeemable on 8 October 2006; and that both coupons are paid half-yearly in arrear on8 April and 8 October. Assuming that the RPI will grow continuously at a rate of 2.5% per annum from its latestknown value,

(i) Show that the real yield from stock (A) as at 9 October 1997 is 4% per annum effective.(ii) By considering the series of future receipts from stock (B), derive a formula for the present value of £100

nominal of that stock.(iii) Calculate the real yield as at 9 October 1997 from stock (B).



25. (i) Describe the characteristics of an index-linked government bond.(ii) On 1 July 2002, the government of a country issued an index-linked bond of term 7 years. Coupons are paid

half-yearly in arrears on 1 January and 1 July each year. The annual nominal coupon is 2%. Interest and capitalpayments are indexed by reference to the value of an inflation index with a time lag of 8 months.You are given the following values of the inflation index:

Date Inflation indexNovember 2001 110.0May 2002 112.3November 2002 113.2May 2003 113.8

The inflation index is assumed to increase continuously at the rate of 21/2% per annum effective from its valuein May 2003. An investor, paying tax at the rate of 20% on coupons only, purchased the stock on 1 July 2003,just after a coupon payment has been made.Calculate the price to this investor such that a real net yield of 3% per annum convertible half-yearly is obtainedand assuming that the investor holds the bond to maturity.


26. On 15 March 1996, the government of a country issued an index-linked bond of term 6 years. Coupons are payablehalf-yearly in arrears, and the annual nominal coupon rate is 3%.Interest and capital payments are indexed by reference to the value of an inflation index with a time lag of 8 months.A tax-exempt investor purchased the stock at £111 per £100 nominal on 16 September 1999, just after the couponpayment had been made.You are given the following values of the inflation index:

Date Inflation IndexJuly 1995 110.5March 1996 112.1July 1999 126.7September 1999 127.4

(i) Calculate the amount of the coupon payment per £100 nominal stock on 15 March 2000.(ii) Calculate the effective real annual yield to the investor on 16 September 1999. You should assume that the

inflation index will increase from its value in September 1999 at the rate of 4% per annum effective.(iii) Without doing any further calculations, explain how your answer to (ii) would alter, if at all, if the inflation

index for July 1995 had been more than 110.5. (Institute/Faculty of Actuaries Examinations, April 2001) [17]

27. On 15 May 1997, the government of a country issued an index-linked bond of term 15 years. Coupons are payablehalf-yearly in arrears, and the annual nominal coupon rate is 4%.Interest and capital payments are indexed by reference to the value of a retail price index with a time lag of 8 months.The retail price index value in September 1996 was 200 and in March 1997 was 206.The issue price of the bond was such that, if the retail price index were to increase at a rate of 7% p.a. fromMarch 1997, a tax exempt purchaser of the bond at the issue date would obtain a real yield of 3% p.a. convert-ible half-yearly.

(i) (a) Derive the formula for the price of the bond at issue to a tax-exempt investor.(b) Show that the issue price of the bond is £111.53%.

(ii) An investor purchases a bond at a price calculated in (i) and holds it to redemption. The retail price indexincreases continuously at 5% per annum from March 1997. A new tax is introduced such that the investor paystax at 40% on any real capital gain, where the real capital gain is the difference between the redemption moneyand the purchase price revalued according to the retail price index to the redemption date. Tax is only due if thereal capital gain is positive.Calculate the real annual yield convertible half-yearly actually obtained by the investor.


28. The shares of a company currently trade at £2.60 each, and the company has just paid a dividend of 12p per share.An investor assumes that dividends will be paid annually in perpetuity and will grow in line with a constant rate ofinflation. The investor estimates the assumed inflation rate from equating the price of the share with the present valueof all estimated gross dividend payments using an effective interest rate of 6% per annum.

(i) Calculate the investor’s estimation of the effective inflation rate per annum based on the above assumptions.(ii) Suppose that the actual inflation rate turns out to be 3% per annum effective over the following 12 years, but

that all the investor’s other assumptions are correct.Calculate the investor’s real rate of return per annum from purchase to sale, if she sold the shares after 12 yearsfor £5 each immediately after a dividend has been paid. You may assume that the investor pays no tax.



29. An investor is interested in purchasing shares in a particular company. The company pays annual dividends, and adividend of 30 pence per share has just been made. Future dividends are expected to grow at the rate of 5% per annumcompound.

(i) Calculate the maximum price per share that the investor should pay to give an effective return of 9% per annum.(ii) Without doing any further calculations, explain whether the maximum price paid will be higher, lower or the

same if:(a) after consulting the managers of the company, the investor increases his estimate of the rate of growth of

future dividends to 6% per annum.(b) as a result of a government announcement, the general level of future price inflation in the economy is now

expected to be 2% per annum higher than previously assumed.(c) general economic uncertainty means that, whilst the investor still estimates future dividends will grow at

5% per annum, he is now much less sure about the accuracy of this assumption.(Institute/Faculty of Actuaries Examinations, April 2013) [4+6=10]

30. An ordinary share pays dividends on each 31 December. A dividend of 35p per share was paid on 31 December 2011.The dividend growth is expected to be 3% in 2012, and a further 5% in 2013. Thereafter, dividends are expected togrow at 6% per annum compound in perpetuity.

(i) Calculate the present value of the dividend stream described above at a rate of interest of 8% per annum effectivefor an investor holding 100 shares on 1 January 2012.

An investor buys 100 shares for £17.20 each on 1 January 2012. He expects to sell the shares for £18 on 1 Jan-uary 2015.

(ii) Calculate the investor’s expected real rate of return. You should assume that dividends grow as expected anduse the following values of the inflation index:

Year 2012 2013 2014 2015Inflation index at start of year: 110.0 112.3 113.2 113.8


31. Mrs Jones invests a sum of money for her retirement which is expected to be in 20 years’ time. The money isinvested in a zero coupon bond which provides a return of 5% per annum effective. At retirement, the individualrequires sufficient money to purchase an annuity certain of £10,000 per annum for 25 years. The annuity will bepaid monthly in arrear and the purchase price will be calculated at a rate of interest of 4% per annum convertiblehalf-yearly.

(i) Calculate the sum of money the individual needs to invest at the beginning of the 20 year period.The index of retail prices has a value of 143 at the beginning of the 20 year period and 340 at the end of the 20 yearperiod.(ii) Calculate the annual effective real return the individual would obtain from the zero coupon bond.

The government introduces a capital gains tax on zero coupon bonds of 25% of the nominal capital gain.(iii) Calculate the net annual effective real return to the investor over the 20 year period before the annuity com-

mences.(iv) Explain why the investor has achieved a negative real rate of return despite capital gains tax only being a tax on

the profits from an investment. (Institute/Faculty of Actuaries Examinations, September 2013) [5+2+3+2=12]

32. A 91-day government bill is purchased for £95 at the time of issue and is redeemed at the maturity date for £100.Over the 91 days, an index of consumer prices rises from 220 to 222.Calculate the effective real rate of return. (Institute/Faculty of Actuaries Examinations, September 2008) [3]

33. A tax advisor is assisting a client in choosing between 3 types of investment. The client pays tax at 40% on incomeand 40% on capital gains.Investment A requires the investment of £1 million and provides an income of £0.1 million per year in arrears for10 years. Income tax is deducted at source. At the end of the 10 years, the investment of £1 million is returned.In investment B, the initial sum of £1 million accumulates at the rate of 10% per annum compound for 10 years.At the end of 10 years, the accumulated value of the investment is returned to the investor after deduction of capitalgains tax.Investment C is identical to investment B except that the initial sum is deemed, for tax purposes, to have increasedin line with the index of consumer prices between the date of the investment and the end of the 10 year period. Theindex of consumer prices is expected to increase by 4% per annum compound over the period.

(i) Calculate the net rate of return from each of the investments.(ii) Explain why the expected rate of return is higher for Investment C than for Investment B and is higher for

Investment B than for Investment A. (Institute/Faculty of Actuaries Examinations, September 2008) [7+3=10]


34. An ordinary share pays annual dividends. The next dividend is expected to be 6p per share and is due in exactly6 months’ time. It is expected that subsequent dividends will grow at a rate of 6% per annum compound and thatinflation will be 4% per annum. The price of the share is 175p and dividends are expected to continue in perpetuity.Calculate the expected effective real rate of return per annum for an investor who purchases the share.


35. (i) State the characteristics of an equity.An investor was considering investing in the shares of a particular company on 1 August 2014. The investor assumedthat the next dividend would be payable in exactly one year and would be equal to 6 pence per share. Thereafter,dividends will grow at a constant rate of 1% per annum and are assumed to be paid in perpetuity. All dividends willbe taxed at a rate of 20%. The investor requires a net rate of return from the shares of 6% per annum effective.(ii) Derive and simplify as far as possible a general formula which will allow you to determine the value of a share

for different values of:• the next expected dividend;• the dividend growth rate;• the required rate of return;• the tax rate.

(iii) Calculate the value of one share to the investor.The company announces some news that makes the shares more risky.(iv) Explain what would happen to the value of the share, using the formula derived in part (ii).The investor bought 1,000 shares on 1 August 2014 for the price calculated in part (iii). He received the dividendof 6 pence on 1 August 2015 and paid the tax due on the dividend. The investor then sold the shares immediatelyfor 120 pence. Capital gains tax was charged on all gains of at a rate of 25%. On 1 August 2014, the index of retailprices was 123. On 1 August 2015, the index of retail prices was 126.(v) Determine the net real return earned by the investor.

(Institute/Faculty of Actuaries Examinations, September 2015, corrected) [4+5+2+3=14]

CHAPTER 6

Interest Rate Problems

1 Spot rates, forward rates and the yield curve1.1 Basic idea. The interest rate offered on investments usually depends on the term or length of the invest-ment. The term structure of interest rates is concerned with the analysis of the dependence of interest rates onthe length of the investment.

1.2 Spot rates and zero rates; zero coupon bonds. A £1 zero coupon bond of term n is an agreement topay £1 at the end of n years; zero coupon means that no interest payments are made. A zero coupon bond isalso called a pure discount bond.

The term structure of interest rates can be assessed from the prices of zero coupon bonds as follows.Let Pt denote the price at issue of a £1 zero coupon bond which matures after a time period of lengtht years. Let yt denote the yield to maturity of this bond—this is another name for the internal rate ofreturn. This yield, yt, is sometimes called the t-year spot rate of interest . Clearly

(1 + yt)tPt = 1 (1.2a)and hence

yt = P−1/tt − 1

Usually, yt 6= yv when t 6= v. The function y : [0,∞)→ R with y(t) = yt is called the yield curve. An exampleof a yield curve is given in figure 1.2a.

maturity year

yiel

d

2000 2005 2010 2015 2020 2025 2030

0.05

0.06

0.07

0.08

Figure 1.2a. A Yield Curve.Yield curves are plotted against the maturity date.Yield curves usually rise gradually because long bondshave higher yields than short bonds.

Note that yt, the t-year spot rate of interest or the t-year zero rate of interest is the effective rate of interestper annum for an investment which lasts for t years.

Recall that δ, the nominal rate of interest compounded continuously or the force of interest, is defined byi = eδ − 1 or δ = log(1 + i) where i is the effective rate of interest per annum.

With continuous compounding, equation (1.2a) becomes

etYtPt = 1 and hence Yt = −1t

logPt (1.2b)

where Yt is the t-year spot force of interest. Clearly, yt = eYt − 1.Thus Yt corresponds to δ, the nominal rate of interest compounded continuously, but now the dependence onthe length of time of the investment is considered.

Every fixed interest investment can be regarded as a combination of zero coupon bonds as the following exam-ple illustrates.



Example 1.2a. Suppose a bond pays a coupon of size x at times 1, 2, . . . , n and has final redemption value C attime n.(a) Express P , the current price of the bond in terms of x, C and P1, P2, . . . , Pn where P1, P2, . . . , Pn are the pricesof £1 zero coupon bonds with different terms.(b) Express the current price of the bond in terms of x, C and y1, y2, . . . , yn where y1, y2, . . . , yn are the correspond-ing yields of the bonds.(c) Show that the gross redemption yield, i, satisfies miny1, y2, . . . , yn ≤ i ≤ maxy1, y2, . . . , yn.

Solution. The bond can be considered as a combination of n zero coupon bonds all with maturity value x and lengths1, 2, . . . , n and one zero coupon bond with length n and maturity value C.Hence the answer to (a) is: P = x(P1 + P2 + · · · + Pn) + CPn and the answer to (b) is

P = x[

1(1 + y1)

+1

(1 + y2)2 + · · · + 1(1 + yn)n

]+

C

(1 + yn)n

(c) This follows immediately fromx

(1 + y1)+

x

(1 + y2)2 + · · · + x

(1 + yn)n+

C

(1 + yn)n=

x

(1 + i)+

x

(1 + i)2 + · · · + x

(1 + i)n+

C

(1 + i)n

Example 1.2b. Suppose we know the spot rate yk for k = 1, 2, . . . , 10. Find an expression for the price of a bondwith redemption value £100, annual coupon rate 8% and maturing in 10 years in terms of y1, y2, . . . , y10.

Solution. The cash flow is as follows:

Time 0 1 2 3 4 5 6 7 8 9 10Cash flow 0 8 8 8 8 8 8 8 8 8 108

The price is10∑k=1

8(1 + yk)k

+100

(1 + y10)10

1.3 Determining the spot rate. There are few zero coupon bonds and so the zero rate must also be calculatedfrom the prices of other financial instruments. Here are two possibilities:

• Determine y1 from the interest rate on one year Treasury bills. To find y2, suppose we observe the price Pof a two year bond with redemption value C, face value f and coupon rate r payable annually. Now the priceof the bond should equal the net present value of the cash flow stream, which is:

Time 1 2Cash flow fr C + fr

Hence

P =fr

1 + y1+fr + C

(1 + y2)2

As y1 is known, the value of y2 can be determined. Then the value of y3 can be determined from the price of athree year bond, and so on.

• A zero coupon bond can also be constructed by subtracting bonds with identical maturity dates but differentcoupons.Suppose bond A is a ten year bond with price PA, coupon rA, face value and redemption value C.Suppose bond B is a ten year bond with price PB , coupon rB , face value and redemption value C.Without loss of generality, assume rA > rB .Consider a portfolio of −rBC of bond A and rAC of bond B. This portfolio will have a face value of(rA − rB)C, a price of P = rAPB − rBPA and zero coupon. Hence

(1 + y10)10P = (rA − rB)C

and y10 can be determined.

6 Interest Rate Problems Sep 27, 2016(9:51) Section 1 Page 103

1.4 Par yield of a bond. The n-year par yield of a bond is the coupon rate that causes the current bond priceto be equal to its face value, assuming the bond to be redeemed at par.

As usual, let f denote the face value, r the coupon rate, P the current price. Assume the coupon is payableannually for n years. The cash flow is as follows:

Time 0 1 2 . . . n− 1 nCash flow −P fr fr . . . fr f + fr

We want P = f . This implies that

the n-year par yield, r, satisfies 1 = rn∑k=1

1(1 + yk)k

+1

(1 + yn)n(1.4a)

where yk is the k-year spot rate of interest.The par yields give an alternative measure of the relationship between the yield and the term of an investment.The difference between the n-year par yield and the n-year spot rate is called the n-year coupon bias. Thus then-year coupon bias is r − yn where r is the n-year par yield.

1.5 Forward interest rates. Suppose fT,k denotes the interest rate per annum for money which will beborrowed at some time T in the future for a period of time of length k. This is called a forward interest rate.Clearly f0,k = yk.

...........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

........

........

........

..

........

........

........

..

........

........

........

.............. ............

0 Tstart of investment

T + kend of investment

time

....................................................................................................................................................................................................................................................................... ....................................period of investment

Now £1 invested at time 0 grows to £(1 + yT+k)T+k at time T + k. Also £1 grows to £(1 + yT )T at time T . Ifthis amount of £(1 + yT )T is reinvested then it grows to £(1 + yT )T (1 + fT,k)r at time T + k. Hence

(1 + yT+k)T+k = (1 + yT )T (1 + fT,k)k

In particular(1 + y2)2 = (1 + y1)(1 + f1,1)

(1 + y3)3 = (1 + y2)2(1 + f2,1) = (1 + y1)(1 + f1,1)(1 + f2,1)

and so on. Hence(1 + yT )T = (1 + f0,1)(1 + f1,1) · · · (1 + fT−1,1) (1.5a)

where f0,1 = y1 is the rate per annum for 1 year starting at time 0; f1,1 is the rate per annum for 1 year startingat time 1, etc.1 Forward rates calculated from equations such as equation (1.5a) are called implied forwardrates as opposed to the market forward rates.

Example 1.5a. Suppose the one year zero rate is 0.07, or 7%, and the two year zero rate is 0.08 or 8%. Find theimplied one year forward rate for a period of length 1 year.

Solution. Recall that the “two year zero rate” is the rate per annum for an investment of length two years. So we aregiven that y1 = 0.07 and y2 = 0.08. Using (1 + y2)2 = (1 + y0)(1 + f1,1) gives f1,1 = 1.082/1.07− 1 = 0.090 or 9%.

1.6 Continuous compounding. For continuous compounding, we havePTPT+k

=[1 + fT,k

]k = ekFT,k

where the forward force of interest is FT,k = ln(1 + fT,k). Hence

FT,k =lnPT − lnPT+k

k(1.6a)

=(T + k)YT+k − TYT

kNote that F0,k = Yk where Yk is defined in equation (1.2b) on page 101.

1 If there is no possibility of confusion, the quantity fk,1 is often abbreviated to fk.


Example 1.6a. Suppose the spot force of interest for an investment of 1 year is a nominal 7% per annum com-pounded continuously and the spot force of interest for an investment of 6 months is a nominal 6% per annumcompounded continuously. Find the implied force of interest for an investment starting in 6 months’ time and lastingfor 6 months.

Solution. The general result ise(t+k)F0,t+k = etF0,t × ekFt,k or (t + k)F0,t+k = tF0,t + kFt,k

In our case, t = k = 0.5, F0,1 = 0.07 and F0,0.5 = 0.06. Hence F0.5,0.5 = 0.08. So the answer is a nominal 8% perannum compounded continuously.

1.7 A more sophisticated model with allowance for interest rates varying with time.• Yields which depend on time and the length of the investment. The above equations are too simplistic:we should really treat £P , the price of a £1 zero coupon bond, as a function of two variables instead of justone. So, for 0 ≤ t ≤ T , let P (t, T ) denote the value at time t of a zero coupon bond with face value 1 andmaturing at time T .Clearly P : (x, y) ∈ R× R : 0 ≤ x ≤ y → R with P (x, x) = 1 for all x ≥ 0.

This function P of two variables will be complicated: for example, the prices of 10 year bonds and 15 yearbonds will tend to move together in the short term. Hence the functions t→ P (t, t + 10) and t→ P (t, t + 15)will be related. fs Corresponding to this function of two variables, we have the yield to maturity, y(t, T ) definedby

[1 + y(t, T )]T−t P (t, T ) = 1 for all 0 ≤ t ≤ T . (1.7a)

Note that y(t, T ) denotes the yield to maturity from a zero coupon bond that starts at time t and ends at time T .

We can now plot the yield curve at time t: this is a plot of the function T → y(t, T ) for T ≥ t. Figure (1.2a)on page 101 is such a plot with t = 2000.

This model reduces to the previous model if we assume the price just depends on the length of time betweenissue and maturity and not on the date of issue. This is equivalent to assuming P (t, T ) = P (0, T − t) for all t.Then set Pt = P (0, t). As P (t, T ) determines Y (t, T ) and conversely, it follows that if P (t, T ) = P (0, T − t)for all t, then y(t, T ) = y(0, T − t) for all t, and conversely. Set yt = y(0, t). And so we arrive at the simplermodel.

With continuous compounding, we have

e(T−t)Y (t,T )P (t, T ) = 1 where eY (t,T ) = 1 + y(t, T ) and so Y (t, T ) = − lnP (t, T )T − t

• Forward rates. A forward rate is an interest rate agreed now for an investment which starts at some time inthe future.

..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

........

........

........

..

........

........

........

..

........

........

........

..

........

........

........

.............. ............

0 ttime of agreement

Tstart of investment

T + kend of investment

time

......................................................................................................................................................................................................................................................................................................................... ....................................period of investment

Let ft,T,k denote the effective interest rate per annum agreed at time t for an investment starting at time T fora period of length k. Hence the investment matures at time T + k. This means that £1 invested at time T willaccumulate to £(1 + ft,T,k)r at time T + k.The discrete time forward rate can be related to the spot rate as follows: suppose we invest £1 at time t forT − t years and agree to invest the accumulated amount at time T for a further k years. Now the accumulatedamount at time T will be £ [1 + y(t, T )]T−t. This amount will accumulate to £ [1 + y(t, T )]T−t

[1 + ft,T,k

]k attime T + k. But £1 invested at time t for T + k − t years will accumulate to £ [1 + y(t, T + k)]T+k−t. Hence

[1 + y(t, T + k)]T+k−t = [1 + y(t, T )]T−t[1 + ft,T,k

]k (1.7b)

and so, by equation (1.7a) [1 + ft,T,k

]k =[1 + y(t, T + k)]T+k−t

[1 + y(t, T )]T−t=

P (t, T )P (t, T + k)


More generally, repeated applications of equation (1.7b) gives:

[1 + y(t, T )]T−t = [1 + y(t, T − 1)]T−1−t [1 + ft,T−1,1]

= (1 + ft,t,1)(1 + ft,t+1,1) · · · (1 + ft,T−1,1) (1.7c)

Note that ft,t,1 = y(t, t+ 1) is the one-year spot interest rate—it is the one-year riskless rate prevailing at time tfor repayment at time t + 1.

Forward rates calculated from equations such as equation (1.7c) are called implied forward rates as opposed tothe market forward rates.

Relation to previous model. Suppose ft,T,k = f0,T,k for all t; i.e. the forward rate just depends on the startand length of the investment and not when the agreement is made. Denote the common value ft,T,k for all tby fT,k; hence fT,k is the forward rate for money which will be borrowed at some time T in the future for theperiod of time of length k. The model then reduces to the simpler model at the start of this chapter.

• Continuous compounding. For continuous compounding, we have

P (t, T )P (t, T + k)

=[1 + ft,T,k

]k = ekFt,T,k

where the forward force of interest is Ft,T,k = ln(1 + ft,T,k). Hence

Ft,T,k =lnP (t, T )− lnP (t, T + k)

k(1.7d)

=(T + k − t)Y (t, T + k)− (T − t)Y (t, T )

k

by equation (1.7a). Note that Ft,t,k = Y (t, t + k).

• Instantaneous forward rates. The instantaneous forward rate at time t for an investment commencing attime T is defined by

Ft,T = limk→0

Ft,T,k

Using equation (1.7d) shows that

Ft,T = limk→0

logP (t, T )− logP (t, T + k)k

= − limk→0

logP (t, T + k)− logP (t, T )k

= − ∂

∂TlogP (t, T ) (1.7e)

= − 1P (t, T )

∂P (t, T )∂T

Conversely, using P (T, T ) = 1 and integrating equation (1.7e) shows that

P (t, T ) = exp(−∫ T

tFt,u du

)

The instantaneous forward rate is a theoretical concept which is useful in modelling bond prices.

Note that any one of the functions P (t, T ), Y (t, T ) or Ft,T determines the other two. The short rate at time t or theinstantaneous rate at time t is defined to be

rt = Y (t, t) = limk→0

Y (t, t + k) = limk→0

Ft,t,k = Ft,t = − ∂ logP (t, T )∂T

∣∣∣∣T=t

where the third equality follows from Ft,t,k = Y (t, t + k). Hence, rt, which is equal to Ft,t, is the instantaneousforward rate at time t for an investment commencing at time t.There is less information in the function rt than in the functions P (t, T ), Y (t, T ) or Ft,T .


term in years

yiel

d

0 5 10 15 20 25 30

0.05

0.06

0.07

0.08

Figure 1.8a. An Increasing Yield Curve.

term in years

yiel

d

0 5 10 15 20 25 30

0.03

0.04

0.05

0.06

0.07

Figure 1.8b. A Decreasing Yield Curve.

term in years

yiel

d

0 5 10 15 20 25 30

0.03

0.04

0.05

0.06

0.07

Figure 1.8c. A Humped Yield Curve.

1.8 Explaining the term structure of interest rates. The three common types of yield curves are as follows.

• Decreasing yield curve. Long term bonds have lower yields than short term bonds.• Increasing yield curve. Long term bonds have higher yields than short term bonds. This is the most commonshape.• Humped yield curve. Long term bonds have lower yields than short term bonds, but the very short termbonds (less than 1 year) have lower yields than 1 year bonds.

There are three popular explanations of the term structure of interest rates:

(1) Expectations Theory. This theory postulates that the relative attraction of long and short term bondsdepends on the expectations of future movements in interest rates. If investors expect interest rates to fall, thenlong term bonds will become more attractive; this will cause the yields on long term bonds to fall and lead toa decreasing yield curve. Similarly, if investors expect interest rates to rise, then long term bonds will becomeless attractive and so lead to an increasing yield curve.

(2) Liquidity Preference. This theory asserts that investors usually prefer short term investments over longterm ones. The reason for this is that investors anticipate they may need to sell their bonds soon and they knowthat long term bonds are more sensitive to interest changes than short term bonds. Hence purchasing a shortterm bond lessens the risk. This implies that longer-term bonds must offer higher yields as an inducement.

6 Interest Rate Problems Sep 27, 2016(9:51) Exercises 2 Page 107

(3) Market Segmentation. Banks are interested in short term bonds because their liabilities are short term.Pension funds are interested in long term bonds because their liabilities are long term. The market segmentationtheory postulates that the term structure arises from the different forces of supply and demand for bonds ofdifferent lengths.

1.9

Summary. The simple model: interest rates constant in time.• yt denotes the t-year spot rate of interest. This is the effective interest rate p.a. for an investmentwhich lasts for t years.• Pt denotes the price in £ at issue of a £1 zero coupon bond which matures after t years. Hence

Pt(1 + yt)t = 1• ft,k denotes the forward rate for an investment starting at time t for a period of k years. Then:

(1 + yt+k)t+k = (1 + yt)t(1 + ft,k)k and (1 + ft,k)k =(1 + yt+k)t+k

(1 + yt)t=

PtPt+k

(1 + yt)t = (1 + f0,1)(1 + f1,1) · · · (1 + ft−1,1)where ft,1, sometimes written as ft, is the 1-year forward rate for an investment starting at time t for aperiod of 1 year.• Ft,k denotes the forward force of interest for an investment starting at time t for k years. Hence

Ft,k = ln(1 + ft,k) =ln(Pt)− ln(Pt+k)

k

AlsoF0,k = ln(1 + yk) and e(t+k)F0,t+k = etF0,t × ekFt,k

• Ft denotes the instantaneous forward rate at time t. By definition

Ft = limk→0

Ft,k = limk→0

ln(1 + ft,k) = limk→0

ln(Pt)− ln(Pt+k)k

= − d

dtlnPt = − 1

P

dP

dt

Conversely

Pt = exp(−∫ t

0Ft du

)The more sophisticated model: interest rates varying with time.• y(t, T ) denotes the effective rate of interest per annum for an investment starting at time t and maturingat time T .• P (t, T ) denotes the price in £ at time t of a £1 zero coupon bond which matures at time T .Hence P (t, T ) [1 + y(t, T )]T−t = 1.• ft,T,k denotes the forward rate per annum at time t of an investment starting at time T and maturingat time T + k.

Explaining the term structure of interest rates: expectations theory, liquidity preference, marketsegmentation.

Par yield of a bond.


1. At time 0 the 1-year spot rate is 8% p.a., the 2-year spot rate is 9% p.a. and the 3-year spot rate is 91/2% p.a. What isthe value of the 2-year forward rate from time 1? (Institute/Faculty of Actuaries Examinations, April 1997) [2]

2. The following n-year spot rates apply at time t = 0:1 year spot rate of interest: 41/2% per annum effective2 year spot rate of interest: 5% per annum effective3 year spot rate of interest: 51/2% per annum effective

Calculate the 2 year forward rate of interest from time t = 1 expressed as an annual effective rate of interest.(Institute/Faculty of Actuaries Examinations, April 2000) [2]


3. The 1-year forward rates for transactions beginning at times t = 0, 1, 2 are fi, wheref0 = 0.06 f1 = 0.065 f2 = 0.07

Find the par yield for a 3-year bond. (Institute/Faculty of Actuaries Examinations, September 1997) [3]

4. The n year forward rate for transactions beginning at time t and maturing at time t + n is denoted as ft,n. You aregiven:

f0,1 = 6.0% per annum; f0,2 = 6.5% per annum; f1,2 = 6.6% per annum.Determine the 3-year par yield. (Institute/Faculty of Actuaries Examinations, September 1998) [3]

5. (i) The one-year forward rate applying at a particular point in time t is defined as ft,1. If ft,1 = 4%, calculate thecontinuous time forward rate Ft,1 applying over the same period of time.

(ii) Define the instantaneous forward rate Ft.(Institute/Faculty of Actuaries Examinations, September 2004) [2+2=4]

6. In a particular bond market, the two-year par yield at time t = 0 is 4.15% and the issue price at time t = 0 of atwo-year fixed interest stock, paying coupons of 8% annually in arrears and redeemed at 98, is £105.40 per £100nominal. Calculate:(a) the one-year spot rate;(b) the two-year spot rate. (Institute/Faculty of Actuaries Examinations, April 2004) [6]

7. At time t = 0 the n-year spot rate of interest is equal to (2.25 + 0.25n)% per annum effective (1 ≤ n ≤ 5).(a) Calculate the 2-year forward rate of interest from time t = 3 expressed as an annual effective rate of interest.(b) Calculate the 4-year par yield.(c) Without explicitly calculating the gross redemption yield of a 4-year bond paying annual coupons of 3.5%,

explain how you would expect this yield to compare with the par yield calculated in (b).(Institute/Faculty of Actuaries Examinations, April 2006, adapted) [7]

8. The following n-year spot rates were observed at time t = 0:1 year spot rate of interest: 4% per annum2 year spot rate of interest: 5% per annum3 year spot rate of interest: 6% per annum4 year spot rate of interest: 7% per annum5 year spot rate of interest: 71/2% per annum6 year spot rate of interest: 8% per annum

(i) Define what is meant by an n-year spot rate of interest.(ii) Calculate the two-year forward rate of interest at time t = 3.

(iii) Using the above n-year spot rates calculate the 6-year par yield at time t = 0.(Institute/Faculty of Actuaries Examinations, April 1998) [2+2+4=8]

9. The force of interest δ(t) is a function of time and at any time, measured in years, is given by the formulaδ(t) = 0.04 + 0.001t

(i) (a) Calculate the accumulated value of a unit sum of money accumulated from time t = 0 to time t = 8.(b) Calculate the accumulated value of a unit sum of money accumulated from time t = 0 to time t = 9.(c) Calculate the accumulated value of a unit sum of money accumulated from time t = 8 to time t = 9.

(ii) Using your results from (i), or otherwise, calculate:(a) The 8-year spot rate of interest from time t = 0 to time t = 8.(b) The 9-year spot rate of interest from time t = 0 to time t = 9.(c) f8,1, where f8,1 is the one-year forward rate of interest from time t = 8.


10. Three bonds paying annual coupons in arrears of 7% and redeemable at 105 per £100 nominal reach their redemptiondates in exactly one, two and three years’ time, respectively. The price of each of the bonds is £98 per £100 nominal.(i) Determine the gross redemption yield of the 3-year bond.(ii) Calculate all possible spot rates implied by the information given.


11. An individual is investing in a market in which a variety of spot rates and forward contracts are available.If at time t = 0 he invests £1,000 for 2 years, he will receive £1,118 at time t = 2. Alternatively, if at time t = 0 heagrees to invest £1,000 at time t = 1 for 2 years, he will receive £1,140 at time t = 3. However, if at time t = 0 heagrees to invest £1,000 at time t = 1 for one year, he will receive £1,058 at time t = 2.(i) Calculate the following rates per annum effective, implied by this data:

(a) The one year spot rate at time t = 0.(b) The two year spot rate at time t = 0.(c) The three year spot rate at time t = 0.

(ii) Calculate the three year par yield at time t = 0 in this market.(Institute/Faculty of Actuaries Examinations, April 2003) [5+3=8]


12. The n-year spot rate of interest, yn is given by:

yn = 0.04 +n

1000for n = 1, 2 and 3.

(i) Calculate the implied one-year forward rates applicable at times t = 1 and t = 2.(ii) Assuming that coupon and capital payments may be discounted using the same discount factors, and that no

arbitrage applies, calculate:(a) The price at time t = 0 per £100 nominal of a bond which pays annual coupons of 3% in arrears and is

redeemed at 110% after 3 years.(b) The 2-year par yield. (Institute/Faculty of Actuaries Examinations, April 2001) [9]

13. The forward rate from time t− 1 to time t has the following values:f0,1 = 4.0%, f1,2 = 4.5%, f2,3 = 4.8%

(i) Assuming no arbitrage, calculate:(a) the price per £100 nominal of a 3-year bond paying an annual coupon in arrears of 5%, redeemed at par in

exactly 3 years, and(b) the gross redemption yield from the bond.

(ii) Explain why the bond with a higher coupon would have a lower gross redemption yield, for the same term toredemption. (Institute/Faculty of Actuaries Examinations, April, 2002) [7+2=9]

14. Bonds paying annual coupons of 6% annually in arrears and redeemable at par will be redeemed in exactly one year,two years and three years respectively. The price of each of the bonds is £96 per £100 nominal.

(i) Determine the gross redemption yield of the 3 year bond.(ii) Determine the discount factors ν(1), ν(2) and ν(3) that the market is using to discount payments due in 1, 2 and

3 years respectively.(iii) Calculate f0,1, f1,2 and f2,3 where ft−1,t is the forward interest rate from time t− 1 to t.


15. For a particular bond market, zero coupon bonds redeemable at par are priced as follows:bonds redeemable in exactly one year are priced at 97;bonds redeemable in exactly two years are priced at 93;bonds redeemable in exactly three years are priced at 88;and bonds redeemable in exactly four years are priced at 83.

(i) Assuming no arbitrage, calculate:(a) the one-year, two-year, three-year and four-year spot interest rates(b) the rate of return from a bond redeemable at par in 4 years’ time that pays a coupon of 4% annually in

arrears.(ii) Explain why the four-year spot rate is greater than the rate of return from a bond redeemable at par in exactly

4 years’ time paying a coupon of 4% annually in arrears.(iii) Explain the shape of the yield curve indicated by the spot rates calculated in (i)(a) using the liquidity preference

theory, if expectations of future short term interest rates are constant.(Institute/Faculty of Actuaries Examinations, September 2004) [8+2+2=12]

16. Suppose ft,r is the forward rate applicable over the period t to t + r and it is the spot rate over the period 0 to t.The gross redemption yield from a one year bond with a 6% annual coupon is 6% per annum effective; the grossredemption yield from a 2 year bond with a 6% annual coupon is 6.3% per annum effective; and the gross redemptionyield from a 3 year bond with a 6% annual coupon is 6.6% per annum effective. All the bonds are redeemed at parand are exactly one year from the next coupon payment.

(i) (a) Calculate i1, i2 and i3 assuming no arbitrage.(b) Calculate f0,1, f1,1 and f2,1 assuming no arbitrage.

(ii) Explain why the forward rates increase more rapidly with term than the spot rates.(Institute/Faculty of Actuaries Examinations, September 2000) [12]

17. (i) (a) Explain what is meant by the “expectations theory” explanation for the shape of the yield curve.(b) Explain how expectations theory can be modified by both “liquidity preference” and “market segmenta-

tion” theories.(ii) Short-term, one-year annual effective interest rates are currently 10%; they are expected to be 9% in one year’s

time, 8% in two year’s time, 7% in three years’ time and to remain at that level thereafter indefinitely.(a) If bond yields over all terms to maturity are assumed only to reflect expectations of future short-term

interest rates, calculate the gross redemption yields from 1-year, 3-year, 5-year and 10-year zero couponbonds.

(b) Draw a rough plot of the yield curve for zero coupon bonds using the data from part (ii)(a). (Graph paperis not required.)

(c) Explain why the gross redemption yield curve for coupon paying bonds will slope down with a less steepgradient than the zero coupon yield curve.



18. In a particular bond market, n-year spot rates per annum can be approximated by the function 0.08 − 0.04e−0.1n.Calculate:

(i) The price per unit nominal of a zero coupon bond with term 9 years.(ii) The 4-year forward rate at time 7 years.

(iii) The 3-year par yield. (Institute/Faculty of Actuaries Examinations, April 2007) [2+3+3=8]

19. The annual effective forward rate applicable over the period t to t + r is defined as ft,r where t and r are measuredin years. We have f0,1 = 4%, f1,1 = 4.25%, f2,1 = 4.5% and f2,2 = 5%. Calculate the following:

(i) f3,1(ii) All possible zero coupon (spot) yields that the above information allows you to calculate.

(iii) The gross redemption yield of a 4-year bond redeemable at par with a 3% coupon payable annually in arrears.(iv) Explain why the gross redemption yield from the 4-year bond is lower than the 1-year forward rate up to

time 4, f3,1. (Institute/Faculty of Actuaries Examinations, September 2007) [1+4+6+2=13]

20. The n-year spot rate of interest, in, is given by in = a− bn for n = 1, 2 and 3, and where a and b are constants.The one-year forward rates applicable at time 0 and at time 1 are 6.1% per annum effective and 6.5% per annumeffective respectively. The 4-year par yield is 7% per annum.Stating any assumptions:

(i) calculate the values of a and b;(ii) calculate the price per £1 nominal at time 0 of a bond which pays annual coupons of 5% in arrear and is

redeemed at 103% after 4 years. (Institute/Faculty of Actuaries Examinations, April 2008) [4+5=9]

21. Three bonds, paying annual coupons in arrears of 6% are redeemable at £105 per £100 nominal and reach theirredemption dates in exactly one, two and three years’ time respectively. The price of each bond is £103 per £100nominal.

(i) Calculate the gross redemption yield of the three-year bond.(ii) Calculate, to 3 decimal places, all possible spot rates implied by the information given, as annual effective rates

of interest.(iii) Calculate, to 3 decimal places, all possible forward rates implied by the information given, as annual effective

rates of interest. (Institute/Faculty of Actuaries Examinations, September 2008) [3+4+4=11]

22. In a particular bond market, n-year spot rates can be approximated by the function 0.06− 0.02e−0.1n.(i) Calculate the gross redemption yield for a 3-year bond which pays coupons of 3% annually in arrear, and is

redeemed at par. Show all workings.(ii) Calculate the 4-year par yield. (Institute/Faculty of Actuaries Examinations, April 2012) [6+3=9]

23. (i) State the characteristics of a Eurobond.(ii) (a) State the characteristics of a certificate of deposit.

(b) Two certificates of deposit issued by a given bank are being traded. A one-month certificate of deposit pro-vides a rate of return of 12% per annum convertible monthly. A two-month certificate of deposit providesa rate of return of 24% per annum convertible monthly.Calculate the forward rate of interest per annum convertible monthly in the second month, assuming noarbitrage. (Institute/Faculty of Actuaries Examinations, September 2012) [4+4=8]

24. Three bonds each paying annual coupons in arrear of 6% and redeemable at £103 per £100 nominal reach theirredemption dates in exactly one, two and three years’ time respectively.The price of each bond is £97 per £100 nominal.

(i) Calculate the gross redemption yield of the 3-year bond.(ii) Calculate the one-year and two-year spot rates implied by the information given.


25. The force of interest, δ(t), is a function of time and at any time t, measured in years, is given by the formulaδ(t) = 0.05 + 0.002t.

(i) Calculate the accumulated value of a unit sum of money:(a) accumulated from time t = 0 to time t = 7;(b) accumulated from time t = 0 to time t = 6;(c) accumulated from time t = 6 to time t = 7.

(ii) Calculate, using your results from part (i) or otherwise:(a) the seven spot rate of interest per annum from time t = 0 to time t = 7;(b) the six spot rate of interest per annum from time t = 0 to time t = 6;(c) f6,1, where f6,1 is the one year forward rate of interest per annum from time t = 6.

(iii) Explain why your answer to part (ii)(c) is higher than your answer to part (ii)(a).(iv) Calculate the present value of an annuity that is paid continuously at a rate of 30e−0.01t+0.001t2

units per annumfrom t = 3 to t = 10. (Institute/Faculty of Actuaries Examinations, September 2013) [5+3+2+5=15]


26. In a particular country, insurance companies are required by regulation to value their liabilities using spot rates ofinterest derived form the government yield curve.Over time t (measured in years), the spot rate of interest is equal to:

i = 0.02t for t ≤ 5.An insurance company in this country has a group of annuity policies which involve making payments of £1m per an-num for 4 years and £2m per annum in the fifth year. All payments are assumed to be paid halfway through the year.

(i) Calculate the value of the company’s liabilities.(ii) Outline two reasons why the spot yield might rise with term to redemption.

(iii) Calculate the forward rate of interest from time t = 3.5 to time t = 4.5.(Institute/Faculty of Actuaries Examinations, April 2015) [3+3+2=8]

27. Three bonds, each paying annual coupons in arrear of 3% and redeemable at £100 per £100 nominal, reach theirredemption dates in exactly one, two and three years time, respectively. The price of each bond is £101 per £100nominal.

(i) Determine the gross redemption yield of the three-year bond.(ii) Calculate the one-year, two-year and three-year spot rates of interest implied by the information given.

(iii) Calculate the one-year forward rate starting from the end of the second year, f2,1.The pattern of spot rates is upward sloping throughout the yield curve.(iv) Explain, with reference to the various theories of the yield curve, why the yield curve might be upward sloping.


3 Vulnerability to interest rate movements3.1 Background. Suppose an institution holds assets with net present value VA and has liabilities with netpresent value VL. Assume that VA ≥ VL at time 0.

If the rate of interest falls, then both VA and VL will increase. If the rate of interest rises, then both VA andVL will fall. Even for bonds with fixed coupons and fixed redemption values, a bondholder will make a gainor loss when the yield changes because the current value of the bond changes. Clearly the institution wants toensure that VA ≥ VL however the interest rate behaves.

The concepts of duration and convexity give measures of the sensitivity of a cash flow to a movement in interestrates. The concept of immunisation is a strategy for immunising a portfolio to changes in interest rates.

3.2 Duration or Macaulay duration or discounted mean term. Consider the following cash flow:

Time 0 t1 t2 . . . tnCash flow −P ct1 ct2 . . . ctn

We assume that the cash flows ctk do not depend on i. Suppose the force of interest is δ and so the annualeffective rate of interest is i = eδ − 1.Clearly

P =n∑k=1

ctke−δtk =

n∑k=1

ctk(1 + i)tk

The duration or Macaulay duration or discounted mean term is defined by

dM (i) = − 1P

dP

dδ=

1P

n∑k=1

tkctk(1 + i)tk (3.2a)

Hence dM (i) is a weighted mean of the values tk where the weights are ctk/(1 + i)tk . Hence the Macaulayduration is the mean term of the cash flows weighted by their present values.Now 1 + i = eδ. Hence

di

dδ= 1 + i and

dP

di=dP

dδ

dδ

di= ν

dP

dδand so dM (i) = −(1 + i)

1P

dP

di(3.2b)

Clearly dM (i) is a measure of the sensitivity of P to the interest rate i.

Example 3.2a. Consider an n-year bond with face value f , redemption valueC and coupon rate r payable annually.(a) Find the Macaulay duration, dM (i).


(b) Consider the special case of a bond which has a redemption value equal to its face value and is selling at par.(c) Find the Macaulay duration of an n-year zero coupon bond with redemption value of £100.

Solution. (a) The cash flow is as follows:

Time 0 1 2 3 . . . n− 1 nCash flow −P fr fr fr . . . fr C + fr

Let ν = 1/(1 + i) and α = fr/C. Then

P =n∑k=1

fr

(1 + i)k+

C

(1 + i)n= Cαan + Cνn (3.2c)

dM (i) =1P

[n∑k=1

kfr

(1 + i)k+

Cn

(1 + i)n

]=α(Ia)n + nνn

αan + νn

=αν(1− νn − n(1− ν)νn) + (1− ν)2nνn

αν(1− ν)(1− νn) + (1− ν)2νn

=1

1− ν− νn nαν + 1− n(1− ν)

αν(1− νn) + (1− ν)νn(3.2d)

= 1 +1i

+n(i− α)− (1 + i)α[(1 + i)n − 1] + i

(3.2e)

after some algebra. Alternatively, use P = fran,i + Cνn. HencedP

dν= fr(1 + 2ν + · · · + nνn−1) + nCνn−1 =

fr(Ia)n + nCνn

ν

and hence

−dPdi

= −dPdν

dν

di= ν2 dP

dν= ν[fr(Ia)n + nCνn

]and so on.(b) In this case, C = P = f ; hence α = r. Substituting these equalities in equation (3.2c) gives α = i. Hence

dM (i) =1P

[n∑k=1

kfrνk + Cnνn]

= in∑k=1

kνk + nνn =1− νn

1− νAlternatively, just substitute α = i = (1− ν)/ν in equation (3.2d).(c) The only payment is £100 at time n. Hence

dM (i) =100nνn

100νn= n

Example 3.2b. Consider an n-year bond with face value f , redemption value C and coupon rate r payable half-yearly. Let i denote the effective annual yield and define i(2) in the usual way by (1 + i(2)/2)2 = 1 + i.(a) Find the Macaulay duration, dM (i).(b) Show that

dM (i) = −(

1 +i(2)

2

)1P

dP

di(2)

Solution. Clearly,

P =fr

2

2n∑k=1

1(1 + i)k/2 +

C

(1 + i)n

Substituting into equation (3.2a) gives

dM (i) =1P

[fr

2

2n∑k=1

k/2(1 + i)k/2 +

Cn

(1 + i)n

]For part (b), note that

di

di(2) = 1 +i(2)

2

Using this result in equation (3.2b) gives

dM (i) = −(1 + i)1P

dP

di= −

(1 +

i(2)

2

)2 1P

dP

di= −

(1 +

i(2)

2

)1P

dP

di(2)


3.3 Effective duration or modified duration or volatility. Suppose i denotes the effective interest rate perannum. Then:

P =n∑k=1

ctk(1 + i)tk

The effective duration, d(i) or volatility or modified duration of the cash flow is defined to be

d(i) = − 1P

dP

di=

1P

n∑k=1

tkctk(1 + i)tk+1 (3.3a)

where the last equality assumes that the values of the cash flows ctj do not depend on i.

Thus the effective duration is also a measure of the rate of change of the net present value with i which doesnot depend on the size of the net present value.

Note thatdM (i) = (1 + i)d(i)

Using the approximation f (x + h) ≈ f (x) + hf ′(x) shows that if interest rates change from i to i + ε thenP (i + ε) ≈

(1− εd(i)

)P (i)

and henceP (i + ε)− P (i)

P (i)≈ −εd(i)

This leads to the following interpretation of the modified duration: if the yield i decreases by ε then the priceincreases by the proportion 100εd(i)%.

Example 3.3a. Suppose the modified duration of a bond is equal to 8. Suppose further that the bond’s yieldincreases by 10 basis points. (A basis point is 0.01%.) Find the approximate change in the price of the bond.

Solution. The approximate change is −εd(i) = −0.001× 8 = −0.008. It decreases by 0.8% approximately.

3.4 Some results about duration.• The Macaulay duration of a zero-coupon bond equals the time to maturity—by part (c) of example (3.2a).• The duration increases as the yield decreases—see exercise 2 in section 4 below.• The duration of a bond increases as the coupon rate decreases (other things being equal)—by equation (3.2e).

Duration is a measure of the riskiness of a bond; it is important for three reasons:• it is a summary measure of the average payments which are due;• it is a measure of the sensitivity of the price to interest rate changes;• it is useful concept for immunising portfolios against interest rate changes.

Here are some standard examples:

Example 3.4a. Find the Macaulay duration and the effective duration of a constant perpetuity:Time 0 1 2 3 . . .Cash flow, c −P (i) 1 1 1 . . .

Solution. Now P (i) = ν + ν2 + ν3 + · · · = ν/(1− ν). Hence:

dM (i) =1P

(ν + 2ν2 + 3ν3 + · · ·) =1P

ν

(1− ν)2 =1

1− ν=

1 + ii

and d(i) =1i

Example 3.4b. Find the Macaulay duration of a constant annuity:Time 0 1 2 · · · n− 1 nCash Flow, c −P (i) 1 1 · · · 1 1

Solution. P (i) = ν + ν2 + · · · + νn and

dM (i) =ν + 2ν2 + · · · + nνn

ν + ν2 + · · · + νn=

11− ν

− nνn

1− νn=

1 + ii− n

(1 + i)n − 1


3.5 Convexity. The convexity of the cash flow above is defined to be

c(i) =1P

d2P

di2=

1P

n∑k=1

tk(tk + 1)ctk(1 + i)tk+2

The duration (first derivative) and convexity (second derivative) together give information about the local de-pendence of P on i; they show how P changes when there is a small change in i. The second order Taylorapproximation is:

f (x + h) = f (x) + hf ′(x) +h2

2f ′′(xu) where xu ∈ (x, x + h)

Applying this to the net present value gives

P (i + ε) ≈ P (i) + εdP

di+ε2

2d2P

di2

and henceP (i + ε)− P (i)

P (i)≈ −εd(i) +

ε2

2c(i)

Convexity measures how fast the slope of a bond’s price curve changes when the interest rate changes—so higher positive convexity is good for the investor. This is illustrated in figure (3.5a): if the yield falls,then the price of bond with higher convexity (CD) will rise more than the price of the bond with smallerconvexity (AB); conversely, if the yield rises, then the price of the bond with higher convexity (CD) will fallless than the price of the bond with smaller convexity (AB). If two bonds have the the same price, the sameyield and the same same duration but different convexities, then the investor should prefer the bond with thehigher convexity—hence they should not have the same price!

i0 yield, i

price, P

P0

A

B

C

D

dPdi at i = i0

Figure 3.5a. Suppose the price of a bond is P0 when the yield is i0.Then higher convexity (CD) is preferable to lower convexity (AB).

(wmf/convexity,0pt,0pt)

Finally, note that if the yield falls by the amount ∆, then the rise in the price is larger than the fall in the priceif the yield increases by the amount ∆.

3.6 Immunisation. Suppose a fund has assets with cash flow atk and net present value VA(i). Let dA(i)denote the effective duration and cA(i) denote the convexity. Suppose further that the fund has liabilities withcash flow `tk and net present value VL(i). Let dL(i) denote the effective duration and cL(i) denote theconvexity.

Suppose that at rate of interest i0 we have VA(i0) = VL(i0). The fund is immunised against small changes ofsize ε in the interest rate if VA(i0 + ε) ≥ VL(i0 + ε).Consider the difference S(i) = VA(i)− VL(i). Taylor’s theorem gives

S(i0 + ε) = S(i0) + εS′(i0) +ε2

2S′′(i0) + · · ·

Using S(i0) = 0 shows that


S(i0 + ε) = εS′(i0) +ε2

2S′′(i0) + · · ·

Dividing by VA(i0) = VL(i0) givesS(i0 + ε)VA(i0)

= εS′(i0)VA(i0)

+ε2

2S′′(i0)VA(i0)

+ · · ·

Now if we insist effective durations are equal at i0, we must have V ′A(i0) = V ′L(i0) and soS′(i0)VA(i0)

= 0

and henceS(i0 + ε)VA(i0)

=ε2

2S′′(i0)VA(i0)

+ · · · = ε2

2[cA(i)− cL(i)] > 0 provided that cA(i0) > cL(i0)

We have shown that if(1) VA(i0) = VL(i0), the net present values are the same(2) dA(i0) = dL(i0), the effective durations are the same(3) cA(i0) > cL(i0), the convexity of the assets is greater than the convexity of the liabilitiesthen we have Redington immunisation at interest rate i0.(Note that condition (3) implies that a decrease in interest rates will cause asset values to increase by more thanthe increase in liability values; similarly an increase in interest rates will cause asset values to decrease by lessthan a decrease in liability values.)

Redington immunisation is not usually a feasible method for organising assets. It will not usually be possibleto keep the effective durations equal; nor will it be possible to assess the future cash flows with any certainty.

3.7 Alternatively, differentiation with respect to the force of interest could be used. Recall equation (3.2b):

dP

dδ= (1 + i)

dP

diwhich implies dM (i) = (1 + i)d(i).

It also leads tod2P

dδ2 =di

dδ

dP

di+ (1 + i)

d

dδ

dP

di= (1 + i)

dP

di+ (1 + i)

di

dδ

d2P

di2

and hence1P

d2P

dδ2 + dM (i) = (1 + i)2c(i)

Hence the three conditions for Redington immunisation are equivalent to:(1) VA(i0) = VL(i0), the net present values are the same(2) dM (A)(i0) = dM (L)(i0), the Macaulay durations are the same(3) cδA(i0) > cδL(i0), the convexity of the assets is greater than the convexity of the liabilities where now theconvexity is cδ = 1

Pd2Pdδ2 .

Note that the three conditions above can be expressed in terms of the original cash flows as follows:

n∑k=1

atk(1 + i)tk

=n∑k=1

ltk(1 + i)tk

n∑k=1

tkatk(1 + i)tk

=n∑k=1

tkltk(1 + i)tk

n∑k=1

t2katk(1 + i)tk

>

n∑k=1

t2kltk(1 + i)tk

This is often the most useful version of the test conditions for numerical problems.


3.8

Summary.• Macaulay duration or discounted mean term. This is the mean term of the cash flows weighted bytheir present values.

dM (i) =1P

n∑k=1

tkctk(1 + i)tk

= − 1P

dP

dδ= −(1 + i)

1P

dP

di

• Effective duration or modified duration or volatility.

d(i) =1P

n∑k=1

tkctk(1 + i)tk+1 = − 1

P

dP

di

• Convexity.

c(i) =1P

d2P

di2=

1P

n∑k=1

tk(tk + 1)ctk(1 + i)tk+2

• Immunisation. Redington immunisation occurs at interest rate i0 if(1) VA(i0) = VL(i0), the net present values are the same(2) dA(i0) = dL(i0), the effective durations are the same(3) cA(i0) > cL(i0), the convexity of the assets is greater than the convexity of the liabilities.Alternative conditions are:

n∑k=1

atk(1 + i)tk

=n∑k=1

ltk(1 + i)tk

n∑k=1

tkatk(1 + i)tk

=n∑k=1

tkltk(1 + i)tk

n∑k=1

t2katk(1 + i)tk

>n∑k=1

t2kltk(1 + i)tk


1. A particular bond has just been issued and pays coupons of 10% per annum paid half yearly in arrears and is redeemedat par after 10 years. Find the duration of the bond in years, at a rate of interest of 5% per half year effective.


2. By equation (3.2a), the Macaulay duration can be regarded as the expectation of a random variable, X , with thefollowing distribution:

X = tk with probabilityctkν

tk∑nk=1 ctkν

tkfor k = 1, 2, . . . , n.

Show that the derivative of dM (i) with respect to i is equal to −νσ2 where σ2 = var[X]. Hence show that theduration increases as the yield decreases2.

3. An insurance company has a continuous payment stream of liabilities to meet over the coming 20 years. The paymentstream will be at a rate of £10 million per annum throughout the period.Calculate the duration of the continuous payment stream at a rate of interest of 4% per annum effective.


4. (a) An investor holds a portfolio of fixed interest securities to meet future liabilities. State the conditions that needto be met if the investor is to be immunised from small, uniform changes in the rate of interest.

(b) State the relationship between “volatility” and “duration” where “volatility” is defined as

− 1A

dA

diwhere A is the value of the portfolio of investments and i is the annual effective rate of interest.

(c) A perpetuity pays annual coupons in arrears. Show that the volatility of the perpetuity, as defined in (b) above,is equal to 1/i where i is the rate of return. (Institute/Faculty of Actuaries Examinations, September 2004) [6]

5. A new management team has just taken over the running of a finance company. They discover that the companyhas liabilities of £15 million due in 13 years’ time and £10 million in 25 years’ time. The assets consist of two zerocoupon bonds, one paying £12.425 million in 12 years’ time and the other paying £12.946 million in 24 years’ time.The current interest rate is 8% per annum effective.Determine whether the necessary conditions are satisfied for the finance company to be immunised against smallchanges in the rate of interest. (Institute/Faculty of Actuaries Examinations, April 2003) [8]

2 See also page 81 of Bond Pricing and Portfolio Analysis by Olivier de La Grandville.


6. (i) State the features of a eurobond.

(ii) An investor purchases a eurobond on the date of issue at a price of £97 per £100 nominal. Coupons are paidannually in arrear. The bond will be redeemed at par 20 years from the issue date. The rate of return from thebond is 5% per annum effective.

(a) Calculate the annual rate of coupon paid by the bond.

(b) Calculate the duration of the bond. (Institute/Faculty of Actuaries Examinations, September 2005) [3+6=9]

7. An insurance company has liabilities of £10 million due in 10 years’ time and £20 million due in 15 years’ time,and assets consisting of two zero-coupon bonds, one paying £7.404 million in 2 years’ time and the other paying£31.834 million in 25 years’ time. The current interest rate is 7% per annum effective.

(i) Show that Redington’s first two conditions for immunisation against small changes in the rate of interest aresatisfied for this insurance company.

(ii) Determine the profit or loss, expressed as a present value, that the insurance company will make if the interestrate increases immediately to 7.5% per annum effective.

(iii) Explain how you might have anticipated, before making the calculation in (ii), whether the result would be aprofit or loss. (Institute/Faculty of Actuaries Examinations, April 2004) [5+2+2=9]

8. An insurance company has liabilities of £87,500 due in 8 years’ time and £157,500 due in 19 years’ time. Its assetsconsist of two zero coupon bonds, one paying £66,850 in 4 years’ time and the other paying £X in n years’ time.The current interest rate is 7% per annum effective.

(i) Calculate the discounted mean term and convexity of the liabilities.

(ii) Determine whether values of £X and n can be found which ensure that the company is immunised against smallchanges in the interest rate. (Institute/Faculty of Actuaries Examinations, April 2007) [5+5=10]

9. A small insurance fund has liabilities of £4 million due in 19 years’ time and £6 million in 21 years’ time. Themanager of the fund has sold the assets previously held and is creating a new portfolio by investing in the zero-coupon bond market. The manager is able to buy zero-coupon bonds for whatever term he requires and has adequatemonies at his disposal.

(i) Explain whether it is possible for the manager to immunise the fund against small changes in the rate of interestby purchasing a single zero-coupon bond.

(ii) In fact, the manager purchases two zero-coupon bonds, one paying £3.43 million in 15 years’ time and the otherpaying £7.12 million in 25 years’ time. The current interest rate is 7% per annum effective.Investigate whether the insurance fund satisfies the necessary conditions to be immunised against small changesin the rate of interest. (Institute/Faculty of Actuaries Examinations, April 2005) [2+8=10]

10. A pension fund has liabilities of £3 million due in 3 years’ time, £5 million due in 5 years’ time, £9 million duein 9 years’ time, and £11 million due in 11 years’ time. The fund holds two investments, X and Y . Investment Xprovides income of £1 million payable at the end of each year for the next 5 years with no capital repayment.Investment Y is a zero-coupon bond which pays a lump sum of £R at the end of n years (where n is not necessarilyan integer). The interest rate is 8% per annum effective.

(i) Investigate whether values of £R and n can be found which ensure that the fund is immunised against smallchanges in the interest rate.You are given that

∑5t=1 t

2νt = 40.275 at 8%.

(ii) (a) The interest rate immediately changes to 3% per annum effective. Calculate the revised present value ofthe assets and liabilities of the fund.

(b) Explain your answer to (ii)(a). (Institute/Faculty of Actuaries Examinations, April 2006) [8+4=12]

11. An insurance company has a portfolio of annuity contracts under which it expects to pay £1 million at the end ofeach of the next 20 years., followed by £0.5 million at the end of each of the following 20 years. The governmentbond with the longest duration in which it can invest its funds pays a coupon of 10% per annum in arrears and isredeemed at par in 15 years’ time. The yield to maturity of the government bond is 6% per annum effective and acoupon payment has just been made.

(i) (a) Calculate the duration of the insurance company’s liabilities at a rate of interest of 6% per annum.(b) Calculate the duration of the insurance company’s assets at a rate of interest of 6% per annum effective, if

all the insurance company’s funds are invested in the government bond with the longest duration.

(ii) (a) Explain why the insurance company cannot immunise its liabilities by purchasing government bonds.(b) Without any further calculations, state the circumstances under which the insurance company would make

a loss if there were a uniform change in interest rates. Explain why a loss would be made.(Institute/Faculty of Actuaries Examinations, September 2003) [8+4=12]


12. (i) An investment provides income of £1 million payable at the end of each year for the next 10 years. There isno capital repayment. If the interest rate is 7% per annum effective, show that the “discounted mean term” (or“Macaulay duration”) of the investment is 4.9646 years.

(ii) An investment company has liabilities of £7 million due in 5 years’ time and £8 million due in 8 years’ time. Thecompany holds two investments,A andB. InvestmentA is the investment described in part (i) and InvestmentBis a zero coupon bond which pays £X at the end of n years (where n is not necessarily an integer).The interest rate is 7% per annum effective. Investigate whether values of £X and n can be found which ensurethat the investment company is immunised against small changes in the interest rate.

You are given that∑10t=1 t

2νt = 228.451 at 7%. (Institute/Faculty of Actuaries Examinations, April 2001) [12]

13. (i) Prove that

(Ia)n =an − nνn

iA government bond pays a coupon half-yearly in arrears of £10 per annum. It is to be redeemed at par in exactly10 years. The gross redemption yield from the bond is 6% per annum convertible half-yearly.

(ii) Calculate the duration of the bond in years.(iii) Explain why the duration of the bond would be longer if the coupon rate were £8 per annum instead of £10 per

annum. (Institute/Faculty of Actuaries Examinations, April, 2002) [3+8+2=13]

14. A fixed interest security was issued on 1 January in a given year. The security pays half-yearly coupons of 4% perannum. The security is redeemable at 110% 20 years after issue. An investor who pays both income tax and capitalgains tax at a rate of 25% buys the security on the date of issue. Income tax is paid on coupons at the end of thecalendar year in which the coupon is received. Capital gains tax is paid immediately on sale or redemption.

(i) Calculate the price paid by the investor to give a net rate of return of 6% per annum effective.(ii) Calculate the duration of the net payments from the fixed interest security for an investor who pays income tax

as described above but who does not pay capital gains tax, at a rate of interest of 6% per annum effective.(Institute/Faculty of Actuaries Examinations, September 2004) [6+8=14]

15. A pension fund has the following liabilities: annuity payments of £160,000 per annum to be paid annually in arrearsfor the next 15 years and a lump sum of £200,000 to be paid in 10 years. It wishes to invest in two fixed interestsecurities in order to immunise its liabilities. SecurityA has a coupon rate of 8% per annum and a term to redemptionof 8 years. Security B has a coupon rate of 3% per annum and a term to redemption of 25 years. Both securities areredeemable at par and pay coupons annually in arrear.

(i) Calculate the present value of the liabilities at a rate of interest of 7% per annum effective.(ii) Calculate the discounted mean term of the liabilities at a rate of interest of 7% per annum effective.

(iii) Calculate the nominal amount of each security that should be purchased so that both the present value anddiscounted mean term of assets and liabilities are equal.

(iv) Without further calculation, comment on whether, if the conditions in (iii) are fulfilled, the pension fund is likelyto be immunised against small, uniform changes in the interest rate.


16. A company incurs a liability to pay £1,000(1 + 0.4t) at the end of year t, for t equal to 5, 10, 15, 20 and 25. Itvalues these liabilities assuming that in the future there will be a constant effective interest rate of 7% per annum.An amount equal to the total present value of the liabilities is immediately invested in 2 stocks:

Stock A pays coupons of 5% per annum annually in arrears and is redeemable in 26 years at par.Stock B pays coupons of 4% per annum annually in arrears and is redeemable in 32 years at par.

The gross redemption yield on both stocks is the same as that used to value the liabilities.(i) Calculate the present value of the liabilities.

(ii) Calculate the discounted mean term of the liabilities.(iii) If the discounted mean term of the assets is the same as the discounted mean term of the liabilities, calculate the

nominal amount of each stock which should be purchased.(Institute/Faculty of Actuaries Examinations, September 2002) [3+3+9=15]

17. An analyst is valuing two companies, Cyber plc and Boring plc. Cyber plc is assumed to pay its first annual dividendin exactly 6 years. It is assumed that the dividend will be 6p per share. After the sixth year, annual dividends areassumed to grow at 10% per annum compound for a further 6 years. Dividends are assumed to grow at 3% per annumcompound in perpetuity thereafter. Boring plc is expected to pay an annual dividend of 4p per share in exactly oneyear. Annual dividends are then expected to grow thereafter by 0.5% per annum compound in perpetuity. The analystvalues dividends from both shares at a rate of interest of 6% per annum effective in a discounted dividend model.

(i) (a) Calculate the value of Cyber plc.(b) Calculate the value of Boring plc.


There is a general rise in interest rates and the analyst decides it is appropriate to increase the valuation rate of interestto 7% per annum effective.(ii) Show that the percentage fall in the value of Cyber plc is greater than that in the value of Boring plc.

(iii) Explain your answer to part (ii) in terms of duration.(Institute/Faculty of Actuaries Examinations, April, 2002) [8+6+2=16]

18. Let Ctk denote a series of cash flows at times tk for k = 1, 2, . . . , n.(i) (a) Define the volatility of the cash flow series, and derive a formula expressing the volatility in terms of tk,

Ctk and ν.(b) Define the convexity of the cash flow series and derive a formula expressing the convexity in terms of tk,

Ctk and ν.(ii) A loan stock issued on 1 March 1997 has coupons payable annually in arrear at 8% p.a. Capital is to be redeemed

at par 10 years from the date of issue.(a) Show that the volatility of this stock at 1 March 1997 at an effective rate of interest of 8% p.a. is 6.71.(b) At 1 March 1997 an investor has a liability of £100,000 to be paid in 7.247 years. Calculate the volatility

and convexity of this liability at 1 March 1997 at an effective rate of interest of 8% p.a.(c) On 1 March 1997 the investor decides to invest a sum equal to the present value of the liability in the

loan stock, where the present value of the liability and the price of the loan stock are both calculated at aneffective rate of interest of 8% p.a.Given that the convexity of the loan stock at 1 March 1997 is 60.53, state with reasons whether the investorwill be immunised against small movements in interest rates on that date.

(iii) Calculate the present value of investor’s profit or loss at 1 March if, immediately after purchasing the loan stockthe rate of interest changes to 81/2% p.a. effective.


19. An investor has to pay a lump sum of £20,000 at the end of 15 years from now and an annuity certain of £5,000 perannum payable half-yearly in advance for 25 years, starting in 10 years’ time.The investor currently holds an amount of cash equal to the present value of these two liabilities valued at an effectiverate of interest of 7% per annum.The investor wishes to immunise her fund against small movements in the rate of interest by investing the cash intwo zero coupon bonds, bond X and bond Y . The market prices of both bonds are calculated at an effective rate ofinterest of 7% per annum. The investor has decided to invest an amount in bond X sufficient to provide a capital sumof £25,000 when bond X is redeemed in 10 years’ time. The remainder of the cash is invested in bond Y .In order to immunise her holdings:

(i) Calculate the amount of money invested in bond Y .(ii) Determine the term needed for bond Y and the redemption amount payable at the maturity date.

(iii) Without doing any further calculations, state which other condition needs to be satisfied for immunisation to beachieved successfully. (Institute/Faculty of Actuaries Examinations, April 1998) [5+10+2=17]

20. (i) Let Ctk denote a series of cash flows at times tk for k = 1, 2, . . . , n and P (i) denote the present value ofthese payments at an effective interest rate i so that P (i) =

∑nk=1 Ctk/(1 + i)tk .

(a) Define the discounted mean term (or Macaulay duration) of the cash flows in terms of of tk, Ctk and ν.(b) Define the volatility (or effective duration) of the cash flows in terms of P (i) and show that for this series

of cash flows the discounted mean term = volatility× (1 + i).(ii) A fund has to provide an annuity of £60,000 per annum payable yearly in arrears for the next 9 years followed

by a final payment of £750,000 in 10 years’ time.The fund has earmarked cash assets exactly equal to the present value of the payments and the fund managerwants to invest these in two zero coupon bonds: Bond A which is repayable at the end of 5 years and Bond Bwhich is repayable at the end of 20 years.The fund manager wants both the assets and the liabilities to have the same volatility. To achieve this, howmuch should the manager invest in each of the bonds, given that an effective rate of interest of 7% per annum isused to value both assets and liabilities?

(iii) State without doing any further calculations, what further condition would be required to ensure that the fund isimmunised against possible losses due to small changes in the effective rate of interest.


21. A pension fund expects to make payments of £100,000 per annum at the end of each of the next 5 years. It wishesto immunise these liabilities by investing in 2 zero coupon bonds which mature in 5 years and in 1 year respectively.The rate of interest is 5% per annum effective.

(i) (a) Show that the present value of the liabilities is £432,948.(b) Show that the duration of the liabilities is 2.9 years.

(ii) Calculate the nominal amounts of the 2 zero coupon bonds which must be purchased if the pension fund is toequate the present value and duration of assets and liabilities.


(iii) (a) Calculate the convexity of the assets.(b) Without calculating the convexity of the liabilities, comment on whether you think Redington’s immunisa-

tion has been achieved. (Institute/Faculty of Actuaries Examinations, September 2000) [18]

22. A variable annuity, payable annually in arrears for n years is such that the payment at the end of year t is t2. Showthat the present value of the annuity can be expressed as

2(Ia)n − an − n2νn+1

1− νA pension fund has a liability to pay £100,000 at the end of one year, £105,000 at the end of 2 years, and so on, theamount increasing by £5,000 each year to £195,000 at the end of 20 years, this being the last payment. The fundvalues these payments using an effective interest rate of 7% per annum. This is also the interest rate at which thecurrent prices of all bonds are calculated.The fund invests an amount equal to the present value of these liabilities in the following two assets:(A) a zero coupon bond redeemable in 25 years, and(B) a fixed interest bond redeemable at par in 12 years’ time which pays a coupon of 8% per annum annually in

arrears.

(a) Calculate the present value and the duration of the liabilities.(b) Calculate the amount of cash that should be invested in each asset if the duration of the assets is to equal that of

the liabilities. (Institute/Faculty of Actuaries Examinations, April 2000) [20]

23. An investor purchases a fixed interest security. The security pays coupons at a rate of 10% p.a. half-yearly in arrear,and is to be redeemed at 110 in 20 years. The investor is subject to tax on the coupon payments at a rate of 25%.

(i) (a) Show that the price paid by the investor to obtain a rate of return of 10% p.a. effective is £81.76%.(b) Calculate the effective duration (or volatility) of the security at 10% p.a. effective for this investor at the

purchase date.

(ii) The investor has two liabilities. The present value of the liabilities, at an effective rate of interest of 10% p.a. isequal to the present value of the investor’s holding in the fixed interest security described above. The amount ofeach of the two liabilities is the same. The second liability is due in 10 years.(a) Show that the first liability must be due in 9.61 years in order that the effective duration (or volatility) of

the liabilities is equal to the effective duration of the assets.(b) Calculate the convexity of the total of the two liabilities described above.(c) Without any further calculations, state with reasons whether the fund comprising the asset and liabilities

described is immunised against small movements in the rate of interest.

(iii) One year after purchasing the fixed interest security, the price at which the security can be sold is still at a levelthat will yield a rate of return of 10% p.a. effective. At this time, a second investor agrees a forward contract tobuy the security four years later, immediately after the coupon payment then due.Calculate the forward price based on a “risk-free” rate of return of 10% p.a. effective and no arbitrage.

(Institute/Faculty of Actuaries Examinations, May 1999, Specimen Examination) [9+12+4=25]

24. A pension fund has liabilities to pay pensions each year for the next 60 years. The pensions paid will be £100m at theend of the first year, £105m at the end of the second year, £110.25m at the end of the third year and so on, increasingby 5% each year. The fund holds government bonds to meet its pension liabilities. The bonds mature in 20 years’time and pay an annual coupon of 4% in arrears.

(i) Calculate the present value of the pension fund’s liabilities at a rate of interest of 3% per annum effective.(ii) Calculate the nominal amount of the bond that the fund needs to hold so that the present value of the assets is

equal to the present value of the liabilities.(iii) Calculate the duration of the liabilities.(iv) Calculate the duration of the assets.(v) Using your calculations in (iii) and (iv), estimate by how much more the value of the liabilities would increase

than the value of the assets if there were a reduction in the rate of interest to 1.5% per annum effective.(Institute/Faculty of Actuaries Examinations, September 2007) [4+3+6+4+4=21]

25. (i) An investor is considering the purchase of an annuity, payable annually in arrear for 20 years. The first paymentis £500. Using a rate of interest of 8% per annum effective, calculate the duration of the annuity when:(a) the payments remain level over the term;(b) the payments increase at a rate of 8% per annum compound.

(ii) Explain why the answer in (i)(b) is higher than the answer in (i)(a).(Institute/Faculty of Actuaries Examinations, April 2008) [6+2=8]

26. An insurance company is considering two possible investment options.

The first investment option involves setting up a branch in a foreign country. This will involve an immediate outlayof £0.25 million, followed by investments of £0.1 million at the end of one year, £0.2 million at the end of two years,£0.3 million at the end of three years and so on until a final investment is made of £1 million in 10 years’ time. The


investment will provide annual payments of £0.5 million for 20 years with the first payment at the end of the 8th year.There will be an additional incoming cash flow of £5 million at the end of the 27th year.

The second investment option involves the purchase of 1 million shares in a bank at a price of £4.20 per share. Theshares are expected to provide a dividend of 21p per share in exactly one year, 22.05p per share in two years and soon, increasing by 5% per annum compound. The shares are expected to be sold at the end of 10 years, just after adividend has been paid, for £5.64 per share.

(i) Determine which of the options has the higher net present value at a rate of interest of 7% per annum effective.(ii) Without doing any further calculations, determine which option has the higher discounted mean term at a rate

of interest of 7% per annum effective. (Institute/Faculty of Actuaries Examinations, September 2008) [9+2=11]

27. A company has a liability of £400,000 due in 10 years’ time.The company has exactly enough funds to cover the liability on the basis of an effective interest rate of 8% per annum.This is also the interest rate on which current market prices are calculated and the interest rate earned on cash.The company wishes to hold 10% of its funds in cash, and to invest the balance in the following securities:• a zero-coupon bond redeemable at par in 12 years’ time• a fixed-interest stock which is redeemable at 110% in 16 years’ time bearing interest at 8% per annum payableannually in arrear.

(i) Calculate the nominal amounts of the zero-coupon bond and the fixed-interest stock which should be purchasedto satisfy Redington’s first two conditions for immunisation.

(ii) Calculate the amount which should be invested in each of the assets mentioned in (i).(iii) Explain whether the company would be immunised against small changes in the rate of interest if the quantities

of stock in part (i) are purchased. (Institute/Faculty of Actuaries Examinations, September 2008) [10+2+2=14]

28. A company has the following liabilities:• annuity payments of £200,000 per annum to be paid annually in arrear for the next 20 years;• a lump sum of £300,000 to be paid in 15 years.The company wishes to invest in two fixed-interest securities in order to immunise its liabilities.

Security A has a coupon rate of 9% per annum and a term to redemption of 12 years.Security B has a coupon rate of 4% per annum and a term to redemption of 30 years.

Both securities are redeemable at par and pay coupons annually in arrear. The rate of interest is 8% per annumeffective.

(i) Calculate the present value of the liabilities.(ii) Calculate the discounted mean term of the liabilities.

(iii) Calculate the nominal amount of each security that should be purchased so that Redington’s first two conditionsfor immunisation against small changes in the rate of interest are satisfied for this company.

(iv) Describe the further calculations that will be necessary to determine whether the company is immunised againstsmall changes in the rate of interest. (Institute/Faculty of Actuaries Examinations, April 2012) [3+4+8+2=17]

29. (i) Describe three theories that have been put forward to explain the shape of the yield curve.

The government of a particular country has just issued five bonds with terms to redemption of one, two, three, fourand five years respectively. The bonds are redeemed at par and have coupon rates of 4% per annum payable annuallyin arrear.

(ii) Calculate the duration of the one year, three year and five year bonds at a gross redemption yield of 5% per an-num effective.

(iii) Explain why a five year bond with a coupon rate of 8% per annum would have a lower duration than a five yearbond with a coupon rate of 4% per annum.

Four years after issue, immediately after the coupon payment then due the government is anticipating problemsservicing its remaining debt. The government offers two options to the holders of the bond with an original term offive years:Option 1: the bond is repaid at 79% of its nominal value at the scheduled time with no final coupon payment beingpaid.Option 2: the redemption of the bond is deferred for 7 years from the original redemption date and the coupon rateis reduced to 1% per annum for the remainder of the existing term and the whole of the extended term.

Assume the bonds were issued at a price of £95 per £100 nominal.(iv) Calculate the effective rate of return per annum from Option 1 and 2 over the total life of the bond and determine

which would provide the higher rate of return.(v) Suggest two other considerations that bond holders may wish to take into account when deciding which options

to accept. (Institute/Faculty of Actuaries Examinations, September 2012) [7+6+2+6+2=23]

30. Two investment projects are being considered.

(i) Explain why comparing the two discounted payback periods or comparing the two payback periods are notgenerally appropriate ways to choose between two investment projects.


The two projects each involve an initial investment of £3 million. The incoming cash flows from the two projects areas follows:Project A. In the first year, Project A generates cash flows of £0.5 million. In the second year, it will generate cashflows of £0.55 million. The cash flows generated by the project will continue to increase by 10% per annum until theend of the sixth year and will then cease. Assume that all cash flows are received in the middle of the year.Project B. Project B generates cash flows of £0.64 million per annum for 6 years. Assume that all cash flows arereceived continuously throughout the year.

(ii) (a) Calculate the payback period from project B.(b) Calculate the discounted payback period from project B at a rate of interest of 4% per annum effective.

(iii) Show that there is at least one “cross-over point” for projects A and B between 0% per annum effective and4% per annum effective where the cross-over point is defined as the rate of interest at which the net presentvalue of the two projects is equal.

(iv) Calculate the duration of the incoming cash flows from projects A and B at a rate of interest of 4% per annumeffective.

(v) Explain why the net present value of projectA appears to fall more rapidly than the net present value of projectBas the rate of interest increases. (Institute/Faculty of Actuaries Examinations, September 2012) [3+5+6+6+2=22]

31. An insurance company has liabilities of £6 million due in 8 years’ time and £11 million due in 15 years’ time. Theassets consist of two zero-coupon bonds, one paying £X in 5 years’ time and the other paying £Y in 20 years’ time.The current interest rate is 8% per annum effective. The insurance company wishes to ensure that it is immunisedagainst small changes in the rate of interest.

(i) Determine the values of £X and £Y such that the first two conditions for Redington’s immunisation are satisfied.(ii) Demonstrate that the third condition for Redington’s immunisation is also satisfied.


32. A pension fund has liabilities to meet annuities payable in arrear for 40 years at a rate of £10 million per annum.

The fund is invested in two fixed interest securities. The first security pays annual coupons of 5% and is redeemedat par in exactly 10 years’ time. The second security pays annual coupons of 10% and is redeemed at par in exactly5 years’ time. The present value of the assets in the pension fund is equal to the present value of the liabilities of thefund and exactly half the assets are invested in each security. All assets and liabilities are valued at a rate of interestof 4% per annum effective.

(i) Calculate the present value of the liabilities of the fund.(ii) Calculate the nominal amount of each security purchased by the pension fund.

(iii) Calculate the duration of the liabilities of the pension fund.(iv) Calculate the duration of the assets of the pension fund.(v) Without further calculations, explain whether the pension fund will make a profit or loss if interest rates fall

uniformly by 1.5% per annum effective.(Institute/Faculty of Actuaries Examinations, September 2013) [1+6+3+4+2=16]

33. A fixed interest security, redeemable at par in 10 years, pays annual coupons of 9% in arrear and has just been issuedat a price to give an investor who does not pay tax a rate of return of 7% per annum effective.

(i) Calculate the price of the security at issue.(ii) Calculate the discounted mean term (duration) of the security at issue.

(iii) Explain how your answer to part (ii) would differ if the annual coupons on the security were 3% instead of 9%.(iv) (a) Calculate the effective duration (volatility) of the security at the time of issue.

(b) Explain the usefulness of effective duration for an investor who expects to sell the security over the nextfew months. (Institute/Faculty of Actuaries Examinations, April 2015) [2+3+2+3=10]

34. An insurance company has sold a pension product to an individual. Under the arrangement, the individual is toreceive an immediate annuity of £500 per year annually in arrear for 12 years. The insurance company has investedthe premium it has received in a fixed-interest bond that pays coupons annually in arrear at the rate of 5% per annumand which is redeemable at par in exactly 8 years.

(i) Calculate the duration of the annuity at an interest rate of 4% per annum effective.(ii) Calculate the duration of the bond at an interest rate of 4% per annum effective.

(iii) State with reasons whether the insurance company will make a profit or a loss if there is a small increase ininterest rates at all terms. (Institute/Faculty of Actuaries Examinations, September 2015) [2+3+2=7]


5 Stochastic interest rate problems5.1 One investment. Suppose £1 is invested at time 0 for n time units. Suppose further that the interest ratesare i1 for time period 1, . . . , in for time period n. Let £Sn denote the accumulated amount at time n. Then

Sn = (1 + i1)(1 + i2) · · · (1 + in)

Now suppose ij is a random variable with mean µj and variance σ2j for j = 1, . . . , n. Suppose further that i1,

. . . , in are independent—this is clearly an unreasonable assumption!!! Then

E[Sn] =n∏j=1

E[1 + ij] =n∏j=1

(1 + µj)

E[S2n] =

n∏j=1

E[1 + 2ij + i2j] =n∏j=1

(1 + 2µj + σ2j + µ2

j)

and an expression for var(Sn) can be obtained by usingvar(Sn) = E[S2

n]− E[Sn]2

If µj = µ and σ2j = σ2 for j = 1, . . . , n, then E[Sn] = (1 + µ)n, E[S2

n] = (1 + 2µ + µ2 + σ2)n and so

var(Sn) = (1 + 2µ + µ2 + σ2)n − (1 + µ)2n

5.2 Repeated investments. Now consider the following sequence of investments:

Time 0 1 2 3 . . . n− 1 nCash flow, c 1 1 1 1 . . . 1 0

Let An denote the value at time n. Then

An =n∑j=1

n∏k=j

(1 + ik) = (1 + i1) · · · (1 + in) + · · · + (1 + in−1)(1 + in) + (1 + in)

and henceAn = (1 + in)(1 +An−1) (5.2a)

where in and An−1 are independent because An−1 depends only on i1, i2, . . . , in−1. Let µn = E[An]. Clearlyµ1 = 1 + µ and E[A2

1] = 1 + 2µ + µ2 + σ2. Taking expectations of equation (5.2a) gives:µn = (1 + µ)(1 + µn−1) for n ≥ 2.

and hence by induction

µn = (1 + µ)n + (1 + µ)n−1 + · · · + (1 + µ) =(1 + µ) [(1 + µ)n − 1]

µ= (1 + µ)sn,µ and sn,µ is tabulated for common n and µ.

Thus E[An] is just the accumulated value sn,µ = (1 + µ)sn,µ calculated at the mean rate of interest.

Calculating the variance of An is straightforward—squaring equation (5.2a) givesA2n = (1 + 2in + i2n)(1 + 2An−1 +A2

n−1)

Using the fact that in is independent of An−1 and taking expectations givesE[A2

n] = (1 + 2µ + µ2 + σ2)(1 + 2µn−1 + E[A2n−1]) for n ≥ 2.

and this gives a recurrence relation for E[A2n]. Then use var[An] = E[A2

n]− µ2n.

Example 5.2a. The yield on a fund is thought to be uniform between 2% and 6%.(a) For a single premium of £1, find the mean and standard deviation of the accumulation after a term of 5 years.(a) Suppose there is an annual premium of £1. Find the mean and standard deviation of the accumulation after a termof 5 years.

Solution. In this case S5 = (1 + i1)(1 + i2)(1 + i3)(1 + i4)(1 + i5) where i1, i2, . . . , i5 are i.i.d. U [0.02, 0.06]. Clearlyµ = E[i] = 0.04 and the density of i is fi(x) = 25 for i ∈ [0.02, 0.06]. The variance of i is given by

σ2 = var[i] =∫ 0.06

0.0225(x− 0.04)2 dx =

25(x− 0.04)3

3

∣∣∣∣0.06

0.02=

430,000


It follows that E[S5] = 1.045 = 1.21665 and E[S25 ] = (1 + 2µ + σ2 + µ2)5 = 1.481157. Hence var[S5]1/2 = 0.03033.

(b) Now µn = E[An] = (1 + µ)sn,0.04 = 1.04sn,0.04. In particular, µ5 = E[A5] = 1.04s5 ,0.04 = 5.63298.Next, using A1 = 1 + i1 gives E[A2

1] = E[(1 + i1)2] = 1 + 2µ + µ2 + σ2 = 1.08173. Denote this quantity by α.Then E[A2

2] = α(1 + 2µ1 + E[A21]) = α(1 + 2.08s1 ,0.04 + E[A2

1]) = 4.5019.Then E[A2

3] = α(1 + 2µ2 + E[A22]) = α(1 + 2.08s2 ,0.04 + E[A2

2]) = 10.5416.Then E[A2

4] = α(1 + 2µ3 + E[A23]) = α(1 + 2.08s3 ,0.04 + E[A2

3]) = 19.5085.Then E[A2

5] = α(1 + 2µ4 + E[A23]) = α(1 + 2.08s4 ,0.04 + E[A2

4]) = 31.7393.

Using var[A5] = E[A25]− E[A5]2 gives var[A5]1/2 = 0.09441.

5.3 The lognormal distribution. Calculating the distribution function of a product of independent randomvariables is usually not possible—however, there is one case where it is possible and that is the case when thefactors have independent lognormal distributions.

Definition. The random variable Y : (Ω,F ,P) → ( (0,∞),B(0,∞) ) has the lognormal distribution withparameters (µ, σ2) iff ln(Y ) has the normal distribution N (µ, σ2).

Here are two memorable properties of the lognormal distribution:

• The product of independent lognormals is lognormal.Suppose Yi ∼ lognormal(µi, σ2

i ) for i = 1, . . . , n and Y1, . . . , Yn are independent. Using the standard resultabout the distribution of the sum of independent normal random variables gives

n∏i=1

Yi = Y1 × · · · × Yn ∼ lognormal(n∑i=1

µi,

n∑i=1

σ2i )

In particular, if Y1, . . . , Yn are i.i.d. lognormal(µ, σ2) then∏Yi ∼ lognormal(nµ, nσ2).

• Mean and variance of a lognormal.If Z ∼ N (µ, σ2) then the moment generating function of Z is

E[etZ] = etµ+ 12 t

2σ2for all t ∈ R

If Y ∼ lognormal(µ, σ2) then Y = eZ where Z ∼ N (µ, σ2). Hence

E[Y ] = E[eZ] = eµ+ 12σ

2and E[Y 2] = E[e2Z] = e2µ+2σ2

and sovar[Y ] = e2µ+σ2

(eσ2 − 1)

Example 5.3a. Suppose £1 is invested at time 0 for n time units. Suppose further that the interest rate is ij for thetime period (j − 1, j) where j = 1, . . . , n and the i1, i2, . . . , in are independent and 1 + ij ∼ lognormal(µj , σ2

j ).(a) Let £Sn denote the accumulated amount at time n. Find the distribution of Sn.(b) Let £Vn denote the present value of £1 due at the end of n years. Find the distribution of Vn.

Solution. (a) Now Sn = (1 + i1)(1 + i2) · · · (1 + in) and lnSn =∑nj=1 ln(1 + ij).

But ln(1 + ij) ∼ N (µj , σ2j ) and the 1 + ij are independent. Hence lnSn ∼ N (

∑nj=1 µj ,

∑nj=1 σ

2j ).

Hence

Sn ∼ lognormal(n∑j=1

µj ,

n∑j=1

σ2j )

(b) Now Vn(1 + i1)(1 + i2) · · · (1 + in) = 1 and hence lnVn = −∑nj=1(1 + ij). Hence

Vn ∼ lognormal(−n∑j=1

µj ,

n∑j=1

σ2j )

Example 5.3b. Suppose Y : (Ω,F ,P)→ ( (0,∞),B(0,∞) ) has the lognormal distribution with parameters (µ, σ2)and E[Y ] = α and var[Y ] = β. Express µ and σ2 in terms of α and β.Solution. We know that α = eµ+ 1

2σ2

and β = e2µ+σ2(eσ

2 − 1). Hence

σ2 = ln(

1 +β

α2

)and eµ =

α√1 + β/α2

=α2√β + α2

or µ = lnα2√β + α2



1. ln(1 + it) is normally distributed with mean 0.06 and standard deviation 0.01. Find Pr[it ≤ 0.05].(Institute/Faculty of Actuaries Examinations, September 1999) [3]

2. The rate of interest in any year has a mean of 6% and standard deviation of 1%. The yield in any year is independentof the yields in all previous years. Find the mean and standard deviation of the accumulated value at time 12 of aninvestment of £10 at time 0.

(Institute/Faculty of Actuaries Examinations, September 1997, adapted) [3]

3. An individual purchases £100,000 nominal of a bond on 1 January 2003 which is redeemable at 105 in 4 years’ timeand pays coupons of 4% per annum at the end of each year.The investment manager wishes to invest the coupon payments on deposit until the bond is redeemed. It is assumedthat the rate of interest at which the coupon payments can be invested is a random variable and the rate of interest inany one year is independent of that in any other year.

Deriving the necessary formulæ, calculate the mean value of the total accumulated investment on 31 December 2006if the annual effective rate of interest has an expected value of 51/2% in 2004, 6% in 2005 and 41/2% in 2006.


4. The yields on an insurance company’s funds in different years are independent and identically distributed. Each yearthe distribution of (1 + i) is lognormal with parameters µ = 0.075 and σ2 = 0.00064, where i is the annual yield onthe company’s funds.

Find the probability that a single investment of £2,000 will accumulate over 10 years to more than £4,500.(Institute/Faculty of Actuaries Examinations, April 1999) [5]

5. An investment bank models the expected performance of its assets over a 5 day period. Over that period, the returnon the bank’s portfolio, i has a mean value of 0.1% and standard deviation 0.2%. Also 1+i is lognormally distributed.

Calculate the value of j such that the probability that i is less then or equal to j is 0.05.(Institute/Faculty of Actuaries Examinations, September 2000) [5]

6. Let it denote the rate of interest earned in the year t− 1 to t. Each year the value of it is 8% with probability 0.625,4% with probability 0.25 and 2% with probability 0.125. In any year, it is independent of the rates of interest earnedin previous years. Let S3 denote the accumulated value of 1 unit for 3 years.

Calculate the mean and standard deviation of S3. (Institute/Faculty of Actuaries Examinations, April 1998) [5]

7. An investment bank models the expected performance of its assets over a 5 day period. Over that period, the returnon the bank’s portfolio, i, has a mean value of 0.15% and standard deviation 0.3%. The quantity 1 + i is lognormallydistributed.

Calculate the value of j such that the probability that i is greater than or equal to j is 0.9.(Institute/Faculty of Actuaries Examinations, September 2003) [6]

8. In any year t, the yield on a fund of investments has mean jt and standard deviation st. In any year, the yield isindependent of the value in any other year. The accumulated value, after n years, of a unit sum of money invested attime 0 is Sn.

(i) Derive formulæ for the mean and variance of Sn if jt = j and st = s for all years t.(ii) (a) Calculate the expected value of S8 if j = 0.06.

(b) Calculate the standard deviation of S8 if j = 0.06 and s = 0.08.(Institute/Faculty of Actuaries Examinations, September 2003) [5+3=8]

9. The expected annual effective rate of return from an insurance company’s investments is 6% and the standard devi-ation of annual returns is 8%. The annual effective returns are independent and (1 + it) is lognormally distributed,where it is the return in the tth year.

(a) Calculate the expected value of an investment of £1 million after 10 years.(b) Calculate the probability that the accumulation of the investment will be less than 90% of the expected value.


10. The rate of interest earned in the year from time t−1 to t is denoted by it. Assume (1+ it) is lognormally distributed.The expected value of the rate of interest is 5%, and the standard deviation is 11%.

(i) Calculate the parameters of the lognormal distribution of (1 + it).

(ii) Calculate the probability that the rate of interest in the year from time t− 1 to t lies between 4% and 7%.(Institute/Faculty of Actuaries Examinations, September 1997) [4+4=8]


11. In any year, the rate of interest on funds invested with a given insurance company is independent of the rates ofinterest in all previous years.Each year the value of 1 + it where it is the rate of interest earned in the tth year, is lognormally distributed. Themean and standard deviation of it are 0.07 and 0.20 respectively.

(i) Determine the parameters µ and σ2 of the lognormal distribution of 1 + it.(ii) (a) Determine the distribution of S15, where S15 denotes the accumulation of one unit of money over 15 years.

(b) Determine the probability that S15 > 2.5.(Institute/Faculty of Actuaries Examinations, April 2000) [5+4=9]

12. The annual rates of interest from a particular investment, in which part of an insurance company’s funds is invested,are independent and identically distributed. Each year, the distribution of (1 + it), where it is the rate of interestearned in year t, is log-normal with parameters µ and σ2.it has mean value 0.07 and standard deviation 0.02, the parameter µ = 0.06748 and σ2 = 0.0003493.

(i) The insurance company has liabilities of £1m to meet in one year from now. It currently has assets of £950,000.Assets can be invested in the risky investment described above or in an investment which has a guaranteedreturn of 5% per annum effective. Find, to two decimal places, the probability that the insurance company willbe unable to meet its liabilities if:(a) All assets are invested in the investment with the guaranteed return.(b) 85% of assets are invested in the investment which does not have the guaranteed return and 15% of assets

are invested in the asset with the guaranteed return.(ii) Determine the variance of return from the portfolios in (i)(a) and (i)(b) above.


13. An insurance company has just written contracts that require it to make payments to policyholders of £1,000,000 in5 years’ time. The total premiums paid by policyholders amounted to £850,000. The insurance company is to investhalf the premium income in fixed interest securities that provide a return of 3% per annum effective. The other halfof the premium income is to be invested in assets that have an uncertain return. The return from these assets in year t,it, has a mean value of 3.5% per annum effective and a standard deviation of 3% per annum effective. The quantities(1 + it) are independently and lognormally distributed.

(i) Deriving all necessary formulæ, calculate the mean and standard deviation of the accumulation of the premiumsover the 5-year period.

(ii) A director of the company suggests that investing all the premiums in the assets with an uncertain return wouldbe preferable because the expected accumulation of the premiums would be greater than the payments due tothe policyholders. Explain why this may be a more risky investment policy.


14. £10,000 is invested in a bank account which pays interest at the end of each year.The rate of interest is fixed randomly at the beginning of each year and remains unchanged until the beginning of thenext year. The rate of interest applicable in any one year is independent of the rate applicable in any other year.During the first year the rate of interest per annum effective will be one of 3%, 4% or 6% with equal probability.During the second year, the rate of interest per annum effective will be either 5% with probability 0.7 or 4% withprobability 0.3.

(i) Assuming that interest is always reinvested in the account, calculate the expected accumulated amount in thebank account at the end of 2 years.

(ii) Calculate the variance of the accumulated amount in the bank account at the end of the 2 years.(Institute/Faculty of Actuaries Examinations, September 2002) [4+7=11]

15. £1,000 is invested for 10 years. In any year, the yield on the investment will be 4% with probability 0.4, 6% withprobability 0.2 and 8% with probability 0.4 and is independent of the yield in any other year.

(i) Calculate the mean accumulation at the end of 10 years.(ii) Calculate the standard deviation of the accumulation at the end of 10 years.

(iii) Without carrying out any further calculations, explain how your answers to parts (i) and (ii) would change (if atall) if:(a) the yields had been 5%, 6% and 7% instead of 4%, 6% and 8% per annum, respectively; or(b) the investment had been made for 12 years instead of 10 years.


16. The annual yields from a particular fund are independent and identically distributed. Each year, the distribution of1 + i is log-normal with parameters µ = 0.07 and σ2 = 0.006, where i denotes the annual yield on the fund.

(i) Find the mean accumulation in 10 years’ time of an investment in the fund of £20,000 at the end of each of thenext 10 years, together with £150,000 invested immediately.

(ii) Find the single amount which should be invested in the fund immediately to give an accumulation of at least£600,000 in 10 years’ time with probability 0.99.



17. In any year the yield on funds invested with a given insurance company has mean value j and standard deviation s,and it is independent of the yields in all previous years.

(i) Derive formula for the mean and the variance of the accumulated value after n years of a single investment of 1at time 0.

(ii) Let it be the rate of interest earned in the tth year. Each year the value of (1 + it) has a lognormal distributionwith parameters µ = 0.08 and σ = 0.04.Calculate the probability that a single investment of £1,000 will accumulate over 16 years to more than £4,250.


18. In any year the yield on a fund of speculative investments has a mean value j and standard deviation s. In any year,the yield is independent of the value in any other year.

(i) Derive formulæ for the mean and variance of the accumulated value after n years, Sn, of a unit investment attime 0.

(ii) Suppose that the random variable (1 + i) where i is the annual yield of the fund in any year, has a lognormaldistribution. Then if the mean value and standard deviation of i are given by j = 6% and s = 1%:(a) Find the parameters µ and σ2 of the lognormal distribution for (1 + i).(b) If S12 is the random variable denoting the accumulation at the end of 12 years of a unit investment show

that S12 has a lognormal distribution with parameters 0.69869 and 0.0010679.(c) Calculate the expected value of S12 and calculate the probability that the accumulation at the end of 12 years

will be at least double the original investment.(Institute/Faculty of Actuaries Examinations, September 1999) [5+8=13]

19. A company is adopting a particular investment strategy such that the expected annual effective rate of return frominvestments is 7% and the standard deviation of annual returns is 9%. Annual returns are independent and (1 + it) islognormally distributed where it is the return in the tth year. The company has received a premium of £1,000 andwill pay the policyholder £1,400 after 10 years.

(i) Calculate the expected value and standard deviation of an investment of £1,000 over 10 years, deriving all theformulæ that you use.

(ii) Calculate the probability that the accumulation of the investment will be less than 50% of its expected value in10 years’ time.

(iii) The company has invested £1,200 to meet its liability in 10 years’ time. Calculate the probability that it willhave insufficient funds to meet its liability.

(Institute/Faculty of Actuaries Examinations, April, 2002) [9+8+3=20]

20. An investment fund provides annual rates of return, which are independent and identically distributed, with annualaccumulation factors following the lognormal distribution with mean 1.04 and variance 0.02.

(i) An investor knows that she will have to make a payment of £5,000 in 5 years’ time.Calculate the amount of cash she should invest now in order that she has a 99% chance of having sufficient cashavailable in 5 years’ time from the investment to meet the payment.

(ii) Comment on your answer to (i). (Institute/Faculty of Actuaries Examinations, April 2004) [9+3=12]

21. (i) In any year, the interest rate per annum effective on monies invested with a given bank has mean value j andstandard deviation s and is independent of the interest rates in all previous years.Let Sn be the accumulated amount after n years of a single investment of 1 at time t = 0.(a) Show that E[Sn] = (1 + j)n.(b) Show that var[Sn] = (1 + 2j + j2 + s2)n − (1 + j)2n.

(ii) The interest rate per annum effective in (i), in any year, is equally likely to be i1 or i2 where i1 > i2. No othervalues are possible.(a) Derive expressions for j and s2 in terms of i1 and i2.(b) The accumulated value at time t = 25 years of £1 million invested with the bank at time t = 0 has expected

value £5.5 million and standard deviation £0.5 million. Calculate the values of i1 and i2.(Institute/Faculty of Actuaries Examinations, April 2005) [5+8=13]

22. An actuarial student has created an interest rate model under which the annual effective rate of interest is assumedto be fixed over the whole of the next 10 years. The annual effective rate is assumed to be 2%, 4% and 7% withprobabilities 0.25, 0.55 and 0.2 respectively.

(a) Calculate the expected accumulated value of an annuity of £800 per annum payable annually in advance overthe next 10 years.

(b) Calculate the probability that the accumulated value will be greater than £10,000.(Institute/Faculty of Actuaries Examinations, April 2006) [4]


23. The rate of interest is a random variable that is distributed with mean 0.07 and variance 0.016 in each of the next10 years. The value taken by the rate of interest in any one year is independent of its value in any other year. Derivingall necessary formulæ, calculate:

(i) The expected accumulation at the end of 10 years, if one unit is invested at the beginning of 10 years.(ii) The variance of the accumulation at the end of 10 years, if one unit is invested at the beginning of 10 years.

(iii) Explain how your answers in (i) and (ii) would differ if 1,000 units had been invested.(Institute/Faculty of Actuaries Examinations, September 2006) [3+5+1=9]

24. £80,000 is invested in a bank account which pays interest at the end of each year. Interest is always reinvested in theaccount. The rate of interest is determined at the beginning of each year and remains unchanged until the beginningof the next year. The rate of interest applicable in any one year is independent of the rate applicable in any other year.During the first year, the annual effective rate of interest will be one of 4%, 6% or 8% with equal probability.During the second year, the annual effective rate of interest will be either 7% with probability 0.75 or 5% with prob-ability 0.25.During the third year, the annual effective rate of interest will be either 6% with probability 0.7 or 4% with probabil-ity 0.3.

(i) Derive the expected accumulated amount in the bank account at the end of 3 years.(ii) Derive the variance of the accumulated amount in the bank account at the end of 3 years.

(iii) Calculate the probability that the accumulated amount in the bank account is more than £97,000 at the end of3 years. (Institute/Faculty of Actuaries Examinations, April 2007) [5+8+3=16]

25. The expected effective annual rate of return from a bank’s investment portfolio is 6% and the standard deviation ofannual effective returns is 8%. The annual effective returns are independent and (1 + it) is lognormally distributed,where it is the return in year t. Deriving any necessary formulæ:

(i) calculate the expected value of an investment of £2 million after 10 years;(ii) calculate the probability that the accumulation of the investment will be less than 80% of the expected value.


26. An insurance company holds a large amount of capital and wishes to distribute some of it to its policyholders by wayof two possible options:Option A. £100 for each policyholder will be put into a fund from which the expected annual effective rate of returnfrom the investments will be 5.5% and the standard deviation of annual returns 7%. The annual effective rates ofreturn will be independent and (1 + it) is lognormally distributed, where it is the rate of return in year t. The policyholder will receive the accumulated investment at the end of ten years.Option B. £100 will be invested for each policyholder for five years at a rate of return of 6% per annum effective.After five years, the accumulated sum will be invested for a further five years at the prevailing five-year spot rate.This spot rate will be 1% per annum effective with probability 0.2, 3% per annum effective with probability 0.3,6% per annum effective with probability 0.2, and 8% per annum effective with probability 0.3. The policyholder willreceive the accumulated investment at the end of ten years.Deriving any necessary formulæ:

(i) Calculate the expected value and the standard deviation of the sum the policyholders will receive at the end ofthe ten years for each of options A and B.

(ii) Determine the probability that the sum the policyholders will receive at the end of ten years will be less than£115 for each of options A and B.

(iii) Comment on the relative risk of the two options from the policyholders’ perspective.(Institute/Faculty of Actuaries Examinations, April 2008) [17+5+2=24]

27. A pension fund holds an asset with current value £1 million. The investment return on the asset in a given year isindependent of returns in all other years. The annual investment return in the next year will be 7% with probability 0.5and 3% with probability 0.5. In the second and subsequent years, annual investment returns will be 2%, 4% and 6%with probability 0.3, 0.4 and 0.3, respectively.

(i) Calculate the expected accumulated value of the asset after 10 years, showing all steps in your calculation.(ii) Calculate the standard deviation of the accumulated value of the asset after 10 years, showing all steps in your

calculation.(iii) Without doing any further calculations, explain how the mean and variance of the accumulation would be

affected if the returns in years 2 to 10 were 1%, 4% or 7% with probability 0.3, 0.4 and 0.3 respectively.(Institute/Faculty of Actuaries Examinations, September 2008) [3+4+2=9]

28. The annual yields from a fund are independent and identically distributed. Each year, the distribution of 1 + i islognormal with parameters µ = 0.05 and σ2 = 0.004, where i denotes the annual yield on the fund.

(i) Calculate the expected accumulation in 20 years’ time of an annual investment in the fund of £5,000 at thebeginning of each of the next 20 years.

(ii) Calculate the probability that the accumulation of a single investment of £1 made now will be greater than itsexpected value in 20 years’ time. (Institute/Faculty of Actuaries Examinations, April 2012) [5+5=10]


29. An individual wishes to make an investment that will pay out £200,000 in 20 years’ time. The interest rate he willearn on the invested funds in the first 10 years will be either 4% per annum with probability of 0.3 or 6% per annumwith probability 0.7. The interest rate he will earn on the invested funds in the second 10 years will also be either4% per annum with probability of 0.3 or 6% per annum with probability 0.7. However, the interest rate in the second10 year period will be independent of that in the first 10 year period.

(i) Calculate the amount the individual should invest if he calculates the investment using the expected annualinterest rate in each 10 year period.

(ii) Calculate the expected value of the investment in excess of £200,000 if the amount calculated in part (i) isinvested.

(iii) Calculate the range of the accumulated amount of the investment assuming the amount calculated in part (i) isinvested. (Institute/Faculty of Actuaries Examinations, September 2012) [2+3+2=7]

30. A cash sum of £10,000 is invested in a fund and held for 15 years. The yield on the investment in any year will be5% with probability 0.2, 7% with probability 0.6 and 9% with probability 0.2, and is independent of the yield in anyother year.

(i) Calculate the mean accumulation at the end of 15 years.(ii) Calculate the standard deviation of the accumulation at the end of 15 years.

(iii) Without carrying out any further calculations, explain how your answers to parts (i) and (ii) would change (if atall) if:(a) the yields had been 6%, 7% and 8% instead of 5%, 7% and 9% per annum, respectively.(b) the investment had been for 13 years instead of 15 years.


31. An insurance company has just written contracts that require it to make payments to policyholders of £10 million infive years’ time. The total premiums paid by policy holders at the outset of the contracts amounted to £7.85 million.The insurance company is to invest the premiums in assets that have an uncertain return. The return from these assetsin year t, it, has a mean value of 5.5% per annum effective and a standard deviation of 4% per annum effective.(1 + it) is independently and lognormally distributed3.

(i) Calculate the mean and standard deviation of the accumulation of the premiums over the five-year period. Youshould derive all necessary formulæ. (Note. You are not required to derive the formulæ for the mean andvariance of a lognormal distribution.)

A director of the insurance company is concerned about the possibility of a considerable loss from the investmentstrategy suggested in part (i). He therefore suggests investing in fixed-interest securities with a guaranteed return of4% per annum effective.(ii) Explain the arguments for and against the director’s suggestion.


32. In any year, the yield on investments with an insurance company has mean j and standard deviation s and is inde-pendent of the yield in all previous years.

(i) Derive formulæ for the mean and variance of the accumulated value after n years of a single investment of 1 attime 0 with the insurance company.

Each year the value of (1 + it), where it is the rate of interest earned in the tth year, is lognormally distributed. Therate of interest has a mean value of 0.04 and standard deviation of 0.12 in all years.(ii) (a) Calculate the parameters µ and σ2 for the lognormal distribution of (1 + it).

(b) Calculate the probability that an investor receives a rate of return between 6% and 8% in any year.(iii) Explain whether your answer to part (ii)(b) looks reasonable.


3 This question is badly phrased. See Mathematical Writing by Donald E. Knuth, Tracy Larrabee and Paul M. Roberts.

CHAPTER 7

Arbitrage

1 Arbitrage

1.1 Arbitrage.

The following is a subset of the definition of arbitrage in the Shorter Oxford English Dictionary:

arbitrage n. & v.A n. Commerce. Trade in bills of exchange or stocks in different markets to take advantage of their differentprices. L19.B v.i. Commerce. Engage in arbitrage. E20.

In finance, an arbitrage opportunity means the possibility of a risk-free trading profit. This could be achievedby making a deal which leads to an immediate profit with no risk of future loss, or making a deal which has norisk of loss but a non-zero probability of making a future profit.

An arbitrageur is a trader who exploits arbitrage possibilities. Such possibilities do not exist for long—becausethey are effectively money-making machines!The term short-selling or shorting means selling an asset that is not owned with the intention of buying itback later. The mechanics are as follows: suppose client A instructs a broker to short sell S. The brokerthen borrows S from the account of another client and sells S and deposits the proceeds in the account of A.Eventually A pays the broker to buy S in the market who returns it to the account he borrowed it from. (It ispossible that the broker will run out of S to borrow and then A will be short-squeezed and be forced to closeout his position prematurely.)

Example 1.1a. Suppose an investor has a portfolio which includes security A and security B. The prices at time 0are as follows: sA = 6 and sB = 11. He assesses the prices at time 1 will be SA1 = 7 and SB1 = 14 if the market goesup and SA1 = 5 and SB1 = 10 if the market goes down.

time 0 time 1 time 1(market rises) (market falls)

A 6 7 5B 11 14 10

Check there is an arbitrage opportunity.

Solution. Suppose the investor sells two A and buys one B at time 0. Hence he makes a gain of 1 at time 0. Attime 1, whether the market rises of falls, there is no change in the value of his portfolio.Clearly, investors will sell A and buy B. This implies the current price of A will fall relative to the price of B, untilsA = sB/2.

Example 1.1b. Suppose an investor has a portfolio which includes security A and security B. The prices at time 0are as follows: sA = 6 and sB = 6. He assesses the prices at time 1 will be SA1 = 7 and SB1 = 7 if the market goes upand SA1 = 5 and SB1 = 4 if the market goes down.

time 0 time 1 time 1(market rises) (market falls)

A 6 7 5B 6 7 4

Check there is an arbitrage opportunity.

Solution. Suppose the investor buys one A and sells one B at time 0. Hence he neither gains nor loses at time 0. Attime 1, there is no change in the value of his portfolio if the market rises; he gains 1 if the market falls.Clearly, investors will buy A and sell B. The current price of A will rise relative to the price of B, until the arbitrageopportunity is eliminated.



1.2 Existence of an arbitrage. Suppose investment A costs the amount sA and investment B costs theamount sB at time t0. Let NPVA(t0) denote the net present value at time t0 of the payoff from investment A.Here are three situations where an arbitrage exists:

• If NPVA(t0) = NPVB(t0) and sA 6= sB then there is an arbitrage.

• If NPVA(t0) 6= NPVB(t0) and sA = sB then there is an arbitrage.

• If sA > sB and NPVA(t0) ≤ NPVB(t0) or if sA < sB and NPVA(t0) ≥ NPVB(t0) then there is anarbitrage.

Example (1.1a) is an example of the first formulation—apply it to two A and one B. Example (1.1b) is anexample of the second formulation. Now consider the third formulation: sA > sB and NPVA(t0) ≤ NPVB(t0).Selling one A and buying one B at time 0 produces a cash surplus at time 0 and increases the value of theportfolio at time t = 1.

1.3 Option pricing.

Example 1.3a. Suppose a share has price 100 at time t = 0. Suppose further that at time t = 1 its price will riseto 150 or fall to 50 with unknown probability. Suppose the effective rate of interest is r per annum.Suppose C denotes the price at time t = 0 of a call option for one share for the exercise price of 125 at the exercisetime of t = 1. Determine C so there is no arbitrage.

Solution. The given information is summarised in the following table:t = 0 t = 1

probability p1 probability 1− p1

share 100 150 50call option at t = 1 for 125 C 25 0

Consider buying x shares and y call options at time t = 0. This will cost 100x + Cy at time t = 0. At time t = 1, thisportfolio will have value 150x + 25y if the share value rises to 150, or 50x if the share value falls to 50.Hence if y = −4x, the value of the portfolio will be 50x whether the share price rises or falls.

Suppose an investor has capital z at time 0 where z > max100, 4C. Consider the following 2 possible decisions:• Investment decision A: buy 1 share, invest the rest;• Investment decision B: buy 4 options, invest the rest.

At time t = 0, the value of the portfolio is z in both cases. For decision A, the value of the portfolio at time t = 1 is

(z − 100)(1 + r) + 150

50= (z − 100)(1 + r) + 50 +

1000

For decision B, the value of his portfolio at time t = 1 is

(z − 4C)(1 + r) + 4× 25 = 100

0For no arbitrage, we can equate these. Hence (100− 4C)(1 + r) = 50 which implies

C =14

(100− 50

1 + r

)We can check this is the condition for no arbitrage as follows:Suppose C > 1

4

(100− 50/(1 + r)

).

At time t = 0, sell 4 options for 4C and borrow 50/(1+r). The total amount received is z = 4C+50/(1+r) > 100.Use 100 to buy one share and the rest is the arbitrage—the rest is the amount z − 100.At time t = 1 sell the share. If the share then has price 150, repay the 50 and use the remaining 100 for payingfor the 4 options; if the share has price 50, then the options are worthless and use the 50 to repay the loan.

Suppose C < 14

(100− 50/(1 + r)

).

At time t = 0, short sell one share for 100; purchase 4 options for 4C and put the rest of the money, which is100− 4C > 50/(1 + r), in the bank.At time t = 1, if the share has price 150, then the 4 options are worth 25 each and there will be more than 50 inthe bank; if the share has price 50, then the options are worthless but there will be more than 50 in the bank. Inboth cases, the obligation of repurchasing the share which was sold short is covered.

More succinctly, comparet = 0 t = 1: probability p1 t = 1: probability 1− p1

share 100 150 50four call options at t = 1 for 125 4C 100 0

If no arbitrage, we must have 4C + 50/(1 + r) = 100.

7 Arbitrage Sep 27, 2016(9:51) Section 2 Page 133

2 Forward contracts

2.1 Historical background. The spot price for a commodity such as copper is the current price for immediatedelivery. This will be determined by the current supply and demand.Suppose a manufacturer requires a supply of copper. He cannot plan his production if he does not know howmuch he will have to pay in 6 months’ time for the copper he then needs. Similarly, the copper supplier doesnot know how to plan his production if he does not know how much he will receive for the copper he producesin 6 months’ time. Both supplier and consumer can reduce the uncertainty by agreeing a price today for the“6 months’ copper”.

2.2 Forward contracts. A forward contract is a legally binding contract to buy/sell an agreed quantity ofan asset at an agreed price at an agreed time in the future. The contract is usually tailor-made between twofinancial institutions or between a financial institution and a client. Thus forward contracts are over-the-counteror OTC. Such contracts are not normally traded on an exchange.Settlement of a forward contract occurs entirely at maturity and then the asset is normally delivered by theseller to the buyer.

2.3 Futures contracts. A futures contract is also a legally binding contract to buy/sell an agreed quantity ofan asset at an agreed price at an agreed time in the future. Futures contracts can be traded on an exchange—thisimplies that futures contracts have standard sizes, delivery dates and, in the case of commodities, quality ofthe underlying asset. The underlying asset could be a financial asset (such as equities, bonds or currencies) orcommodities (such as metals, wool, live cattle).

Suppose the contract specifies that A will buy the asset S from B at the price K at the future time T . ThenB is said to hold a short forward position and A is said to hold a long forward position. Therefore, a futurescontract says that the short side must deliver to the long side the asset S for the exchange delivery settlementprice (EDSP).

On maturity, physical delivery of the underlying asset is not normally made—the short side closes the contractwith the long side by making a cash settlement .

Let ST denote the actual price of the asset at time T .If K > ST then the short side B makes the profit K − ST .If K < ST then the long side A makes the profit ST −K.

With futures contracts, there is daily settlement—this process is called mark to market. At the time the contractis made, the long side is required to place a deposit called the initial margin in a margin account. At the endof each trading day, the balance in the margin account is calculated. If the amount falls below the maintenancemargin then the long side must top-up the account; the long side is also allowed to withdraw any excess overthe initial margin. Generally, interest is earned on the balance in the margin account. Similarly, the short sidealso has a margin account.

We assume in this chapter that there is no difference between the pricing of forward contracts and futurescontracts.

Example 2.3a. Suppose the date is May 1, 2000 and the current spot price of copper is £1,000 per tonne1. Thisis the current price of copper for immediate delivery. Suppose the current price of September copper is £1,100 pertonne. Suppose further that the standard futures contract for copper is for 25 tonnes. This means that the currentprice of a September copper futures contract is 25× £1,100 = £27,500.

Suppose that on May 1, A buys one September futures contract for copper from B. This means that A agrees onMay 1 to pay £27,500 to B for 25 tonnes of copper on September 1 whilst B agrees to sell 25 tonnes of copper to Aon September 1 for £27,500.

1 One tonne is 1,000 kilograms. For current information, see the web site of the London Metal Exchange: www.lme.com.


2.4 Pricing a forward contract with no income. Suppose a forward contract specifies that A will buy theasset S from B at the price K at the future time T . So A is the long side and B is the short side.

Let ST denote the actual price of the asset at time T .If K > ST then the short side B makes the profit K − ST .If K < ST then the long side A makes the profit ST −K.

Assume that the force of interest available on a risk-free investment in (0, T ) is δ. By assuming the no arbitrageassumption, we can calculate the price K.Method I.

Consider the following two investment portfolios:Portfolio A. At time 0:• buy one unit of S at the current price S0.Value of portfolio at time 0: S0. Value of portfolio at time T: one unit of S which has value ST .Portfolio B. At time 0:• enter forward contract to buy one S with forward price K at time T ;• invest Ke−δT in a risk-free investment at force of interest δ.Value of portfolio at time 0: Ke−δT . Value of portfolio at time T: one unit of S which has value ST . (TheamountK from the risk-free investment pays for the forward contract. The size of the investment at time 0is chosen to be Ke−δT in order to ensure that values of the two portfolios are equal at time T .)The no arbitrage assumption implies that

Ke−δT = S0

and henceK = S0e

δT

Method II.We can simplify Method I as follows: equate the time 0 values. Suppose A buys one S at time 0 for theprice S0. At time T , he then has the asset S which he values at K. Equating the time 0 values gives:

S0 = Ke−δT

Note that it has not been necessary to make any assumptions about how the price, St, of the asset varies over(0, T ). The forward price will change over time. Suppose Kt denotes the forward price for a contract taken outat time t for maturity time T . Then Kt = Ste

δ(T−t) where St is the price of the security at time t. As t → Twe have Kt = Ste

δ(T−t) → ST . Hence as t approaches the settlement time, T , the forward price tends to theprice of the underlying asset.

Example 2.4a. The price of a stock is currently 300p. (a) What is the price of a futures contract on this stock witha settlement day in 180 days’ time if the risk-free interest rate is 5% per annum. (Assume there are no dividendspayable over the next 180 days.) (b) After 50 days, suppose the price of the stock has risen to 308p and the risk-freeinterest rate has not changed. Calculate the change in the margin account for an investor holding the futures contractspecified in (a).

Solution. (a) Equating the time 0 values gives S0 = Ke−δt with S0 = 300, t = 180/365 and eδ = 1.05. HenceK = 300 × 1.05180/365 = 307.31. (b) New price of futures contract is Kt = 308 × 1.05130/365 = 313.40. Hence,change in margin account is the gain 313.40− 307.31 = 6.09.

2.5 Pricing a forward contract with a fixed income. Now suppose the security S is a bond which pays thecoupon c at time t1 where t1 ∈ (0, T ).

Example 2.5a. First consider the following simplification: suppose the interest rate is 0% and hence £1 tomorrowis worth exactly £1 today.(a) Suppose the current price of an asset S is £10. It pays a coupon of £1 at time 1. If a forward contract is topurchase S at time 2 for the amount £K, then how much should K be?(b) If the current price of S is S0 and the coupon is c, then how much should K be?

Solution. (a) Here is the justification that K should be £9.Suppose K > 9: then an arbitrageur will buy S now for £10 and enter into a forward contract to sell the asset Sat time 2 for £K. He will receive £1 at time 1 and £K at time 2; hence he makes a risk-free profit of £(K − 9) forevery 1 unit of S.Suppose K < 9. Then an arbitrageur should sell S short now for £10 and enter into a long forward contract to buy Sat time 2 for £K. Of course he must now pay the coupon of £1 at time 1 (which goes to the person from whom S was


borrowed.) and £K at time 2. Hence for every unit of S he makes a profit of £(10− 1−K) = £(9−K). Similarly,anyone owning the asset S at time 0 should make use of the same arbitrage possibility.(b) By a similar argument, K = S0 − c.

Suppose the security S is a bond which pays the coupon c at time t1 where t1 ∈ (0, T ). Suppose a forwardcontract specifies that investor A will buy the asset S from investor B at the price K at the future time T . Wewish to decide what is the appropriate value for K under the assumption that the force of interest available ona risk-free investment in (0, T ) is δ (this means that if we invest an amount b for a length of time h during theinterval (0, T ) then it grows to beδh).Method I.

Consider the following two investment portfolios:Portfolio A. At time 0:• buy one unit of S at the current price S0.Value of portfolio at time 0: S0. Value of portfolio at time T: one unit of S plus amount ceδ(T−t1) (recallthe asset S pays coupon c at time t1 which is then invested).Portfolio B. At time 0:• enter forward contract to buy one S with forward price K maturing at time T ;• invest Ke−δT + ce−δt1 in a risk-free investment at force of interest δ.Value of portfolio at time 0: Ke−δT + ce−δt1 . Value of portfolio at time T: one unit of S plus amountceδ(T−t1). (The amount Ke−δT + ce−δt1 accumulates to K + ceδ(T−t1) and the amountK from the risk-freeinvestment pays for the forward contract. The size of the investment at time 0 is chosen to be Ke−δT +ce−δt1 in order to ensure that values of the two portfolios are equal at time T .)The no arbitrage assumption implies that

Ke−δT + ce−δt1 = S0and hence K = S0e

δT − ceδ(T−t1).Method II.

Equate the time 0 values. Suppose A buys one S at time 0 for the price S0. In return he gets the coupon cat time t1 and has the asset S at time T which he values at K. Equating the time 0 values gives:

S0 = ce−δt1 +Ke−δT

Note that it has not been necessary to make any assumptions about how the price, St, of the asset varies over(0, T ).

Now suppose we have more than one coupon payment: suppose we have coupons of size ck at times tk fork = 1, 2, . . . , n. Then the argument of equating time 0 values gives

Ke−δT = S0 −n∑k=1

cke−δtk

Example 2.5b. The price of a stock is currently 250p. A dividend payment of 10p is expected in 35 days’ timetogether with another dividend payment of 15p after a further 180 days. What is the price of a futures contract onthis stock with a settlement date of 250 days’ time if the risk-free interest rate is 7% per annum.

Solution. We have S0 = 250, T = 250/365 and eδ = 1 + i = 1.07. Equating time 0 values gives

Ke−δT = S0 −n∑k=1

cke−δtk = 250− 10× 1.07−35/365 − 15× 1.07−215/365

leading to K = 236.35.

Example 2.5c. A bond has a current price of £600. Its face value is £1,000 and it matures in exactly 5 years andpays coupons of £30 every 6 months. The first payment is due in exactly 6 months.The 6-month and 12-month interest rates are nominal 6% and 7% per annum compounded continuously.By using the “no arbitrage” assumption, find the price of a forward contract to purchase the bond in exactly one year.

Solution. Let K denote the forward price. Equating time 0 prices gives: 600 = Ke−0.07 + 30e−0.03 + 30e−0.07.

If we wish to equate time T = 1 values then we must argue as follows. We are given the force of interest ratesF0,0.5 = 0.06 and F0,1 = 0.07. Hence the forward force of interest F0.5,0.5 = 0.08 (see example(1.6a) on page 104).This means that the forward force of interest for an investment starting in 6 months’ time is a nominal 8% per annumcompounded continuously. The coupon received in 6 months’ time can be invested and hence will grow to 30e0.04


by time 1. Hence 600e0.07 = 30e0.04 + 30 + K. This is exactly the same equation as the previous one. Both giveK = 582.28.

The price of K = 582.28 can be justified as follows. If K > 582.28, then an investor should buy S now and enterinto a forward contract to sell S for K at time 1 and a contract to invest the dividend at time 0.5 for 6 months.His cash flow is summarised in the following table.

time 0 0.5 1cash flow -600 30 30+K

The NPV is positive.If K < 582.28, then an investor with S should sell S now and enter into a forward contract to buy S for K at time 1.His cash flow is

time 0 0.5 1cash flow 600 -30 -30-K

The NPV is positive.

2.6 Pricing a forward contract with a known dividend yield. Now suppose the security S pays a knowndividend yield which is a nominal r% per annum compounded continuously—the value r is the average overthe life of the forward contract. This means that the dividend payment is proportional to the stock price atthe time of the dividend payment. Hence the actual size of the dividend payments over (0, T ) are unknown attime 0 because the value of the stock price over the time interval (0, T ) is unknown at time 0. We can finessethis problem by assuming that all dividend payments are reinvested in further units of S; hence we assume that1 unit of the security grows to erT units of the security over the period (0, T ).

Portfolio A. At time 0:• enter forward contract to buy one S with forward price K maturing at time T ;• invest Ke−δT in a risk-free investment at force of interest δ.Value of portfolio at time 0: Ke−δT . Value of portfolio at time T: one unit of S. (The amount Ke−δT

accumulates to K which pays for the forward contract.)Portfolio B. At time 0:• buy e−rT units of S at the current price S0 (and reinvest dividend income in further units of S).Value of portfolio at time 0: e−rTS0. Value of portfolio at time T: one unit of S.The no arbitrage assumption implies that

Ke−δT = S0e−rT and hence K = S0e

(δ−r)T

Example 2.6a. The price of a stock is currently 250p. It is expected to provide a dividend income of 2% of the assetprice which is received continuously during a six month period. What is the price of a forward contract on this stockwith a settlement date of 6 months’ time if the risk-free interest rate is a nominal 7% per annum with continuouscompounding.

Solution. Now S0 = 250, δ = 0.07 and er = 1.022. The time 0 value of one unit of stock at time T is S0e−rT .

Equating time 0 values gives Ke−δT = S0e−rT which leads to K = 250e(0.07−r)0.5 = 250e0.035/1.02 = 253.83.

2.7 Value of a forward contract at intermediate times.

Suppose a forward contract is agreed at time 0 with forward price K0 for one unit of S at time T . At time T ,the value of the contract to the long side is ST − K0, the value of the stock at time T minus the price of theforward contract agreed at time 0. Suppose we wish to estimate Vt, the value of the forward contract to the longside at any time t ∈ [0, T ]. Then VT = ST −K0, V0 = 0 and the value to the short side at any time t ∈ [0, T ]is (−Vt).Consider the following 2 portfolios:

Portfolio A. At time t• buy the forward contract agreed at time 0 for the price Vt.• invest K0e

−δ(T−t) in a risk-free investment at force of interest δ for the interval (t, T ).Value of portfolio at time t: Vt + K0e

−δ(T−t). Value of portfolio at time T : one unit of S. (Because theamount K0e

−δ(T−t) grows to K0 which is the amount needed by the forward contract which returns oneunit of S.)Portfolio B. At time t:• agree to a new forward contract with price Kt for one unit of S at time T ;


• invest Kte−δ(T−t) in a risk-free investment at force of interest δ for the interval (t, T ).

Value of portfolio at time t: Kte−δ(T−t). Value of portfolio at time T : one unit of S.

HenceVt +K0e

−δ(T−t) = Kte−δ(T−t)

and so Vt = (Kt −K0)e−δ(T−t). Using Kt = Steδ(T−t) gives Vt = St − S0eδt.

Here is a more succinct derivation: now the long side agrees at time 0 to payK0 at time T for one unit of S andalso agrees at time t to pay Kt at time T for one unit of S. Hence the value to the long side of such a forwardcontract at time t is Kt −K0 which is in time T values. Its value in time t values is Vt = (Kt −K0)e−δ(T−t).Using K0 = S0e

δT and Kt = Steδ(T−t) gives Vt = St − S0eδt.

Example 2.7a. Suppose t0 < t1 < t2 < t3 < t4. Suppose further that the price of a stock at time t0 was P0 andwas P1 at time t1. The stock will pay coupons of C2 at time t2 and C3 at time t3. An investor entered into a longforward contract for 1 unit of the stock at time t0 with the contract due to mature at time t4. Assuming an effectiverate of interest of 5% p.a. and no arbitrage, what is the value of the contract at time t1.

Solution. Let K0 denote the price of the original contract agreed at time t0 and let ν = 1/1.05.Equating the time t0 values gives an equation for K0: K0ν

t4 = P0 − C2νt2 − C3ν

t3 .Let K1 denote the price of a contract agreed at time t1 for 1 unit of the stock to be delivered at time t4.Equating the time t1 values gives an equation for K1: K1ν

t4−t1 = P1 − C2νt2−t1 − C3ν

t3−t1 .Hence the value of the original contract to the long side at time t1 is the difference in prices of these two contracts intime t1 values; and that is (K1 −K0)νt4−t1 = P1 − P0/ν

t1 .

2.8 Speculation, hedging, gearing and leverage.

Speculation is the process of taking a risk in the hope of making a profit.

Example 2.8a. This example is a continuation of example (2.3a).Suppose A buys one September futures contract for copper from B. Then A will have to make a deposit, called themargin. This may be as low as 1% for a financial futures contract. Assume it is 10% for our copper futures contract.Hence A must deposit 10%× £27,500 = £2,750. Suppose that by September, the price of copper has risen to £1,300per tonne. Hence A makes a profit of 25 × (£13,000 − £11,000) = £5,000 for an outlay of £2,750. But note that Ahas taken the risk that he might lose £27,500 for an outlay of £2,750.

Suppose A believes the price of copper will fall. Hence he sells a September futures contract for 25 tonnes of copperfor £27,500. If the price of copper is only £900 per tonne on September 1, then he makes a profit of 25 × £200 =£5,000. But suppose he is wrong and the price of copper doubles to £2,200 per tonne on September 1. Then A loses25× £1,100 = £27,500. Indeed, his potential loss is unlimited.

Gearing and leverage. These terms mean that a large gain or a large loss can be made for a small outlay.Leverage is the term used in the USA whilst gearing is the term used in the UK.

Example 2.8b. The copper example again: examples (2.3a) and (2.8a).Suppose A believes the price will rise. If he trades in the actual product and purchases 25 tonnes of copper today atthe current spot price of £1,000 per tonne, then this will cost him £25,000. If the price rises to £1,300 per tonne onSeptember 1, then he makes a profit of 25× £300 = £7,500 from an investment of £25,000; this profit is 30% of hisinitial investment. His largest possible loss, if the copper becomes worthless, is £25,000 which is 100% of his initialinvestment.Suppose instead of buying the actual product today, he buys a futures contract for 25 tonnes for £27,500. He thenmakes a profit of £5,000 for an outlay of the deposit of £2,750; this profit is 182% of his initial investment. Of course,he is also exposed to a potential loss of £27,500 which is 1,000% of his initial investment!

Hedging is the avoidance of risk. This implies making an investment to reduce the risk of adverse pricemovements in an asset.

Example 2.8c. A UK exporter has won a contract to supply machinery to the US. The exporter will receive apayment of $1m in six month’s time when the machinery is delivered. In order to remove the risk that an adversemovement in the exchange rate makes the deal unprofitable, the exporter purchases a currency futures contract. Thisguarantees that he can exchange the dollars which he will receive in six months’ time into sterling at a rate which isspecified today. In this way, the exporter ensures that the risk that the export deal is made unprofitable by movementsin the exchange rate is removed.

Another example: if you owned a stock, then short selling an equal amount would be a “perfect hedge”. Theterm static hedge means that the hedge portfolio does not change over the period of the contract.


Example 2.8d. Consider again the forward contract made at time 0 to buy one S with forward price K maturing attime T . Find a static hedge for the short side of this simple forward contract. Suppose the risk free force of interestis δ.

Solution. Recall that the short side is the party who contracts to sell one S at time T for the price K. If the short sidejust purchases one unit of S at time 0 and the price falls, then more will have been paid for S then was necessary; ifthe short side just purchases one unit of S at time T to fulfil the contract and the price ST at time T is greater then K(the amount he receives at time T ), then a loss will be made—indeed, this potential loss is theoretically unlimited.

Suppose at time 0 the short side borrows the amount Ke−δT = S0 at the risk-free force of interest δ and buys oneunit of S at the current price S0. The price of this hedge portfolio at time 0 is −Ke−δT + S0 = 0.At time T , the short side has a debt which has grown to K which now equals the amount received under the termsof the forward contract; in addition he owns one unit of S which must be delivered to the long side under the termsof the forward contract. Hence this hedge portfolio means the short side is certain not to make a loss or a profitwhatever the price of the stock at time T .


1. The price of a given share is 80p. The risk-free rate of interest is 5% per annum convertible quarterly. Assumingno arbitrage and that the share will not pay any income, calculate the forward price for the share, for settlement inexactly one quarter of a year. (Institute/Faculty of Actuaries Examinations, April, 2002) [2]

2. A 10-month forward contract is issued on 1 March 2002 on a stock with a price of £12 per share at that date.Dividends of £1.50 per share are expected on 1 June, 1 September and 1 December 2002.Calculate the forward price, assuming a risk-free rate of interest of 6% per annum convertible half-yearly and noarbitrage. (Institute/Faculty of Actuaries Examinations, September 2002) [4]

3. (i) Explain what is meant by the “no arbitrage” assumption in financial mathematics.

(ii) A 7 month forward contract is issued on 1 January 2000 on a stock with a price of £60 per share. Dividends of£2 per share are expected after 3 and 6 months.Assuming a risk-free force of interest of 7% per annum and no arbitrage, calculate the forward price.


4. An asset has a current price of £1.20. Given a risk-free rate of interest of 5% per annum effective and assuming noarbitrage, calculate the forward price to be paid in 91 days.


5. (i) State what is meant by a “forward contract”. Your answer should include reference to the terms “short forwardposition” and “long forward position”.

(ii) A 3-month forward contract is issued on 1 February 2001 on a stock with a price of £150 per share. Dividendsare received continuously and the dividend yield is 3% per annum. In addition, it is anticipated that a specialdividend of £30 per share will be paid on 1 April 2001.Assuming a risk-free force of interest of 5% per annum and no arbitrage, calculate the forward price per shareof the contract. (Institute/Faculty of Actuaries Examinations, April 2001) [6]

6. A one year forward contract is issued on 1 April 2003 on a share with a price of 600p at that date. Dividends of 30pper share are expected on 30 September 2003 and 31 March 2004. The 6-month and 12-month spot risk-free interestrates are 4% and 4.5% per annum effective respectively on 1 April 2003.

Calculate the forward price at issue, assuming no arbitrage.(Institute/Faculty of Actuaries Examinations, September 2003) [4]

7. (i) Explain what is meant by a “forward contract”. Your answer should include reference to the terms “shortforward position” and “long forward position”.

(ii) An investor entered into a long forward contract for £100 nominal of a security seven years ago and the contractis due to mature in 3 years’ time. The price per £100 nominal of the security was £96 seven years ago and isnow £148. The risk-free rate of interest can be assumed to be 4% per annum effective during the contract.Calculate the value of the contract now if the security will pay a single coupon of £7 in two years’ time and thiswas known from the outset. You should assume no arbitrage.


7 Arbitrage Sep 27, 2016(9:51) Exercises 3 Page 139

8. (i) Explain what is meant by the “no arbitrage” assumption in financial mathematics.(ii) A 3-year forward contract is to be issued on a particular company share. The current market value of the share

is £4.50 and a dividend of £0.20 per share has just been paid. The parties to the contract assume that the futurequarterly dividends will increase by 1% per quarter year compound for the first two years and by 11/2% perquarter year compound for the final year.Assuming a risk-free force of interest of 5% per annum and no arbitrage, calculate the forward price.


9. The risk-free force of interest δ(t) at time t is given by:

δ(t) = 0.05 if 0 < t ≤ 10;

0.08 + 0.003t if t > 10.(i) (a) Calculate the accumulation at time 15 of £100 invested at time t = 5.

(b) Calculate the accumulation at time 14 of £100 invested at time t = 5.(c) Calculate the accumulation at time 15 of £100 invested at time t = 14.(d) Calculate the equivalent constant force of interest from time t = 5 to time t = 15.

(ii) Calculate the present value at time t = 0 of a continuous payment stream payable at a rate of 100e0.01t fromtime t = 0 to time t = 5.

(iii) A one year forward contract is issued at time t = 0 on a share with a price of 300p at that date. A dividend of7p per share is expected at time t = 1/2. Calculate the forward price of the share, assuming no arbitrage.


10. A bond is priced at £95 per £100 nominal, has a coupon of 5% per annum payable half-yearly, and has an outstandingterm of 5 years.An investor holds a short position in a forward contract on £1 million nominal of this bond, with a delivery price of£98 per £100 nominal and maturity in exactly one year, immediately following the coupon payment then due.The continuously compounded risk-free rates of interest for terms of 6 months and one year are 4.6% per annum and5.2% per annum, respectively.Calculate the value of this forward contract to the investor assuming no arbitrage.


11. (i) Explain what is meant by the “expectations theory” for the shape of the yield curve.(ii) Short-term, one-year annual effective interest rates are currently 8%; they are expected to be 7% in one year’s

time, 6% in two years’ time and 5% in three years’ time.(a) Calculate the gross redemption yields (spot rates of interest) from 1-year, 2-year, 3-year and 4-year zero

coupon bonds assuming the expectations theory explanation of the yield curve holds.(b) The price of a coupon paying bond is calculated by discounting individual payments from the bond at the

zero coupon bond yields in (a). Calculate the gross redemption yield of a bond that is redeemed at par inexactly 4 years and pays a coupon of 5 per annum annually in arrear.

(c) A two-year forward contract has just been issued on a share with a price of 400p. A dividend of 4p isexpected in exactly in exactly one year.Calculate the forward price using the above spot rates of interest assuming no arbitrage.


12. A share currently trades at £10 and will pay a dividend of 50p in one month’s time. A six-month forward contract isavailable on the share for £9.70. Show that an investor can make a risk-free profit if the risk-free force of interest is3% per annum. (Institute/Faculty of Actuaries Examinations, April 2006) [4]

13. An investor is able to purchase or sell two specially designed risk-free securities, A and B. Short sales of bothsecurities are possible. Security A has a market price of 20p. In the event that a particular stock market index goesup over the next year, it will pay 25p and, in the event that the stock market index goes down, it will pay 15p.Security B has a market price of 15p. In the event that the stock market index goes up over the next year, it willpay 20p and, in the event that the stock market index goes down it will pay 12p.

(i) Explain what is meant by the assumption of “no arbitrage” used in the pricing of derivatives contracts.(ii) Find the market price of B such that there are no arbitrage opportunities and assuming the price of A remains

fixed. Explain your reasoning. (Institute/Faculty of Actuaries Examinations, September 2006) [2+2=4]

14. An investor entered into a long forward contract for a security 5 years ago and the contract is due to mature in 7 years’time. The price of the security was £95 five years ago and is now £145. The risk-free rate of interest can be assumedto be 3% per annum throughout the 12 year period.Assuming no arbitrage, calculate the value of the contract now if

(i) The security will pay dividends of £5 in two years’ time and £6 in four years’ time.(ii) The security has paid and will continue to pay annually in arrear a dividend of 2% per annum of the market

price of the security at the time of payment. (Institute/Faculty of Actuaries Examinations, April 2007) [3+3=6]


15. A one-year forward contract is issued on 1 April 2007 on a share with a price of 900p at that date. Dividends of 50pper share are expected on 30 September 2007 and 31 March 2008. The 6-month and 12-month spot, risk-free ratesof interest are 5% and 6% per annum effective respectively on 1 April 2007.

Calculate the forward price at issue, stating any assumptions.(Institute/Faculty of Actuaries Examinations, September 2007) [4]

16. An 11-month forward contract is issued on 1 March 2008 on a stock with a price of £10 per share at that date.Dividends of 50 pence per share are expected to be paid on 1 April and 1 October 2008.

Calculate the forward price at issue, assuming a risk-free rate of interest of 5% per annum effective and no arbitrage.(Institute/Faculty of Actuaries Examinations, April 2008) [4]

17. (i) A forward contract with a settlement date at time T is issued based on an underlying asset with a current marketprice of B. The annualised risk free force of interest applying over the term of the forward contract is δ and theunderlying asset pays no income. Show that the theoretical forward price is given by K = Beδt, assuming noarbitrage.

(ii) An asset has a current market price of 200p, and will pay an income of 10p in exactly 3 months’ time. Cal-culate the price of a forward contract to be settled in exactly 6 months, assuming a risk free rate of interest of8% per annum convertible quarterly. (Institute/Faculty of Actuaries Examinations, September 2008) [3+3=6]

18. (i) Explain what is meant by the “no arbitrage” assumption in financial mathematics.

An investor entered into a long forward contract for a security four years ago and the contract is due to mature in fiveyears’ time. The price of the security was £7.20 four years ago and is now £10.45. The risk-free rate of interest canbe assumed to be 2.5% per annum effective throughout the nine-year period.

(ii) Calculate, assuming no arbitrage, the value of the contract now if the security will pay dividends of £1.20annually in arrear until maturity of the contract.

(iii) Calculate, assuming no arbitrage, the value of the contract now if the security has paid and will continue to payannually in arrear a dividend equal to 3% of the market price of the security at the time of the payment.


19. A 10 month forward contract was issued on 1 September 2012 for a share with a price of £10 at that date. Dividendsof £1 per share are expected on 1 December 2012, 1 March 2013 and 1 June 2013.

(i) Calculate the forward price assuming a risk-free rate of interest of 8% per annum convertible half yearly and noarbitrage.

(ii) Explain why it is not necessary to use the expected price of the share at the time the forward matures in thecalculation of the forward price. (Institute/Faculty of Actuaries Examinations, September 2012) [4+2=6]

20. (i) Explain the main difference:(a) between options and futures.(b) between call options and put options.

(ii) A one-year forward contract is issued on 1 April 2013 on a share with a price at that date of £10.50. Dividendsof £1.10 per share are expected on 30 September 2013 and 31 March 2014. On 1 April 2013, the 6-month risk-free spot rate of interest is 4.5% per annum convertible half-yearly and the 12-month risk-free rate of interest is5% per annum convertible half yearly.

Calculate the forward price at issue, stating any further assumptions you make.(Institute/Faculty of Actuaries Examinations, April 2013) [4+4=8]

21. A nine-month forward contract is issued on 1 March 2012 on a share with a price of £1.80 at that date. Dividends of10p per share are expected on 1 September 2012. Calculate the forward price at issue assuming a risk-free rate of in-terest of 4% per annum effective and no arbitrage. (Institute/Faculty of Actuaries Examinations, September 2013) [3]

22. (i) Describe what is meant by the “no arbitrage” assumption in financial mathematics.

A 9-month forward contract is issued on 1 April 2015 on a stock with a price of £6 per share on that date. Dividendsare assumed to be received continuously and the dividend yield is 3.5% per annum.

(ii) Calculate the theoretical forward price per share of the contract, assuming no arbitrage and a risk-free force ofinterest of 9% per annum.

(iii) The actual forward price per share of the contract is £6.30 and the risk-free force of interest is as in part (ii).Outline how an investor could make an arbitrage profit.


7 Arbitrage Sep 27, 2016(9:51) Exercises 3 Page 141

23. A 9 month forward contract was issued on 1 October 2015 on a share with a price at that date of £10. Dividendsof 50 pence per share are expected on 1 November 2015 and 1 May 2016. The risk-free force of interest is 5% perannum.

(i) Calculate the forward price at issue, stating any further assumptions made and showing all workings.(ii) Explain why the expected price of the share nine months after issue does not have to be taken into account when

pricing the forward. (Institute/Faculty of Actuaries Examinations, September 2015) [4+2=6]

APPENDIX 1

Some Mathematical PrerequisitesA1.1 Calculus

We know that if f (x) = xn where x > 0 and n ∈ R, then f ′(x) = nxn−1.

Now suppose f (x) = ax where a > 0 is a constant. Then f (x) = ex ln a. Hence f ′(x) = (ln a)ex ln a = (ln a)ax.

An alternative approach: taking logs of f (x) = ax gives ln f (x) = x ln a. Then differentiate to get1

f (x)f ′(x) = ln a

and so f ′(x) = (ln a)ax. The corresponding integration result is displayed below—equation(A1.2a).

A1.2

Summary. If f (x) = ax where a > 0 is a constant, then

f ′(x) = (ln a) ax∫ax dx =

ax

ln a+ constant (A1.2a)

A1.3 Linear Interpolation

Suppose f : R → R is a continuous function. Suppose further that a and b are real numbers with a < b andthe numerical values of f (a) and f (b) are known. We require the value of f (c) where a < c < b and it is notpossible to calculate the value of f (c) directly.

Example 0.3a. The table of the standard normal distribution function, Φ, gives Φ(1.02) = 0.8461 and Φ(1.03) =0.8485. What is the value of Φ(1.025)?

Solution. It seems sensible to give the value which is half way between Φ(1.02) = 0.8461 and Φ(1.03) = 0.8485 andthat is 0.8473.

We can find an approximate value for f (c) by approximating f with a straight line. If the two values of a andb are close and f is continuous, then this approximation should be reasonable.

We are now assuming f is a straight line; hence its gradient is constant.This implies

f (c)− f (a)c− a

=f (b)− f (a)b− a

and hence

f (c) = f (a) +c− ab− a [f (b)− f (a)] (A1.3a)

Using the gradient after c instead of before c gives the following alternative formulation:f (b)− f (c)b− c

=f (b)− f (a)b− a

and hence

f (c) = f (b)− b− cb− a [f (b)− f (a)] (A1.3b)

Both approaches will, of course, give the same approximation.

Example 0.3b. Find the value of x with Φ(x) = 0.5095.

ST334 Actuarial Methods c©R.J. Reed Sep 27, 2016(9:51) Some Mathematical Prerequisites Page 143

Mathematical Prerequisites Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed

Solution. From tables we have Φ(0.02) = 0.5080 and Φ(0.03) = 0.5120. The problem can be expressed in the formof the general solution given above if we set f = Φ−1.

We have f (0.5080) = 0.02 and f (0.5120) = 0.03 and we require the value of f (0.5095). Using equation (A1.3a)with a = 0.5080 and b = 0.5120 gives

f (0.5095) = 0.02 +0.00150.0040

× 0.01 = 0.02375

Using equation (A1.3b) gives

f (0.5095) = 0.03− 0.00250.0040

× 0.01 = 0.02375

However, it is better to argue from first principles as follows. We know Φ(0.02) = 0.5080 and Φ(0.03) = 0.5120 andwe require x with Φ(x) = 0.5095. Using linear interpolation implies

Φ(x)− Φ(0.02)Φ(0.03)− Φ(0.02)

=x− 0.02

0.03− 0.02and hence

x = 0.02 + 0.01× Φ(x)− Φ(0.02)Φ(0.03)− Φ(0.02)

= 0.02 + 0.01× 0.5095− 0.50800.5120− 0.5080

which gives the same answer as before.Note that the actual answer is 0.02382.

APPENDIX 2

AnswersAnswers to Exercises: Chapter 1 Section 3 on page 8 (exs1-1.tex)

1. (a) Using c1 = c0(1 + ni) gives ni = 0.2 with n = 2/12. Hence i = 1.2 or 120% p.a. (b) Using c1 = c0(1 + ni)with c1 = 5100, c0 = 5000 and i = 0.06 gives n = 1/3 or 4 months.

2. Under ACT/360 basis, the return over n days is c0 × 0.095 × n/360. Hence if i denotes the yield under ACT/365,then c0 × 0.095× n/360 = c0 × i× n/365. Hence i = 365× 0.095/360 = 0.0963 or 9.63%.

3. 1000× 1.0112 = 1126.82503 and so the answer to (a) is £126.82 assuming the issuer of the investment rounds in itsfavour, which in this case means down. Also, 1000× 1.0124 = 1269.7346 and so the answer to (b) is £142.91.

4.[

(1 + 0.1)1/12 − 1]× 12 = 0.0957 or 9.57%.

5. The amount is 1000× 1.0112 = 1126.83 assuming the lender rounds in its favour, which in this case means up.6. Denote the effective rate by i. Using(

1 + 0.1× 91365

)= (1 + i)91/365 gives i =

(1 + 0.1× 91

365

)365/91

− 1 = 0.10382

or 10.38%.7. £1000× 1.065/2 = £1156.828. Let i denote the effective annual interest rate. Then 975(1 + i)1/2 = 1000. Hence i = 79/1521 = 0.05194 or 5.19%.9. (a) 4% p.a. payable every 6 months. (b) (1 + 0.02)2 − 1 = 0.0404 or 4.04%. (c) Using (1 + iep)4 = 1.022 givesiep = 1.021/2 − 1 = 0.00995 or 1.00% to two decimal places.

10. (a) It is equivalent to an effective rate of 3% per 6 months which is equivalent to an effective rate of 1.032−1 = 0.0609p.a. So answer is 6.09% p.a. (b) Equivalent to an effective rate of 1.061/2 − 1 = 0.029563 for 6 months. Equivalentto a nominal interest rate of 2× 0.02963 = 0.059126, or 5.91% p.a. payable every 6 months.

11. Equivalent to an effective interest rate of 2% per quarter, which is equivalent to an effective interest rate of 1.022−1,or 4.04% per 6 months, which is equivalent to nominal 8.08% p.a. payable every 6 months.

12. (a) Using 1.013 = 1.0303, 1.016 = 1.0615, 1.0112 = 1.12682 and 1.0124 = 1.2697 gives (i) 3.03%; (ii) 6.15%;(iii) 12.68%; (iv) 26.97%.(b) (i) 12%; (ii) 4× 3.03 = 12.12%; (iii) 12.30%; (iv) 12.68%; (v) 13.485%.

13. The equation

1,500(

1 +91i1365

)= 1540 leads to i1 =

401,500

× 36591

=145

1,365= 0.1069597 and this is 10.70%.

Using

(1 + i2)91/365 =1,5401,500

leads to i2 =(

7775

)365/91

= 0.111331 and this is 11.13%.

14. (a) The effective interest rate is 0.025 every 6 months. Let i = 1.025 and let x be the desired answer. Hence:

x =300

1.0253 +150

1.0257 +500

1.0259 = 805.1338 or £805.13.

(b) The effective interest rate per month is i1 = 1.0251/6 − 1. We want the smallest t with950

(1 + i1)t>

300(1 + i1)24 +

150(1 + i1)48 +

500(1 + i1)60 =

3001.0254 +

1501.0258 +

5001.02510

where the present time is used as the focal point. This gives (1+ i1)t < 1.209427 or t < 46.2032. So required answeris after 46 months.

15. The initial £50,000 leads to payments of £2,500 at the ends of years 1, 2 and 3. Hence we have:Year Assets at start of year Interest at end of year

1 £50,000 at 5% p.a. £2,500£50,000 at 5% p.a. £2,500£2,500 at 6% p.a. £1502

total = £2,650

£50,000 at 5% p.a. £2,500£2,500 at 6% p.a. £150£2,650 at 5.5% p.a. £145.75

3

total = £2,795.75

Overall the amount returned at the end of year 3 is £50,000 + £2,500 + £2,650 + £2,795.75 = £57,945.75.

ST334 Actuarial Methods c©R.J. Reed Sep 27, 2016(9:51) Answers Page 145

Answers 1.3 Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed

16. (a) The interest rate is an effective 1.5% per 3 months. Hence the discount factor for 6 months is ν1 = 1/1.0152

and the discount factor for 1 year is ν = ν21 = 1/1.0154. Denote the original loan by ` and the value of the annual

payments by y. Then

` =800ν1(1− ν20

1 )1− ν1

=yν(1− ν10)

1− ν=yν2

1 (1− ν201 )

1− ν21

Hence y = 800(1 + ν1)/ν1 = 800(1 + 1.0152) = 1624.18. (b) y = x[1 + (1 + i)2

].

17. (a) £2,000/1.073 = £1632.60 (b) £2,000/(1 + 0.07× 182/365) = £1,932.55

18. Denote the answer by i. Then 0.45/360 = i/365. Hence i = 0.45625 or 4.56% p.a.

19. (a) 600 = 500(1 + i)4. Hence 500(1 + i)6 = 500× (6/5)3/2 = 657.27.(b) (i) 5,750 = 5,500(1 + i) and so 1 + i = 23/22. Hence 5,500(1 + i)2 = 6011.37 (ii) 5500 × (1 + i)9/12 = 5686.45.However, a lender may use a simple interest calculation in this situation: this gives 5500× (1 + 9i/12) = 5687.50.

20. (a) The amount c0 grows to c0erk/365 after k days. Hence the equivalent simple annual interest rate is

365k

(erk/365 − 1)

(b) Invert the previous relation to get (365/k) ln(1 + ki/365).

21. (a) (i) Using (1 +

i(m)

m

)m= 1 + i = e0.12

with m = 365/7 gives i(m) = 3657 (e0.12×7/365 − 1) = 0.1201 or 12.01% per annum compounded every 7 days.

(ii) m = 12. Hence i(m) = 12(e0.12/12 − 1) = 0.1206 or nominal 12.06% per annum compounded monthly.(b) (i) £1000e0.12×7/365 = £1002.30; (ii) £1000e0.12/12 = £1010.05.

22. For t ∈ (1, x) we have 1 < t1/n < x1/n. Dividing by t and then integrating this inequality over t between 1 and xgives: ∫ x

1

dt

t<

∫ x

1

dt

t1−1/n < x1/n∫ x

1

dt

t

Hence lnx < n(x1/n − 1) < x1/n lnx. Now x > 0; hence x1/n → 1 as n→∞. Hence result.

23. If i = 0 then f (m) = 0 for all m; hence result.Otherwise, f ′(m) = (1+ i)1/m

[1− ln(1 + i)/m

]−1 and f ′′(m) = (1+ i)1/m

(ln(1+ i)

)2/m3 > 0 for allm ∈ (0,∞).

Hence the function f ′ is strictly increasing. Now limm→∞ f ′(m) = 0. Hence f ′(m) < 0 for all m ∈ (0,∞). Hencef is a decreasing function on (0,∞).

24. A(p) = 1+pip gives the second equality. Using limp↓0 ip = limm↑∞ i(m) = limm↑∞m[(1 + i)1/m − 1

]= ln(1+i) = δ

by using question 22.

25. Define g : [0,∞)→ R by g(x) = (1 + x)t − 1− tx. Suppose t > 1.Then g(0) = 0 and g′(x) = t

[(1 + x)t−1 − 1

]> 0 for all x > 0. Hence for t > 1 and i > 0 we have (1 + i)t > 1 + ti.

Similarly for 0 < t < 1 we have g′(x) < 0 for x > 0; hence for 0 < t < 1 and i > 0 we have (1 + i)t < 1 + ti.

26. (a) 1,000 × 1.1422/365 = 1,116.4949121 or £1,116.49. (b) Interpolate between A(1) = 1000 × 1.1 = 1100 andA(2) = 1000 × 1.12 = 1210. Now A(2) − A(1) = 110. We want A(1) + [A(2)−A(1)] 57/365 = 1,117.18.(c) Because 1 + it > (1 + i)t for 0 < t < 1 by question 25.

27. First investor: £91 grows to £93.90p in 30 days. Let i1 denote the effective interest rate per annum. Then (1 +i1)30/365 = 93.90/91. Hence i1 = 0.4647.Second investor: £93.90 grows to £100 in 60 days. Let i2 denote the effective interest rate per annum. Then(1 + i2)60/365 = 100/93.90. Hence i2 = 0.4665. Hence second investor achieved a higher rate of return.

28. (a) Expected rate of interest is 0.3×0.07+0.5×0.08+0.2×0.1 = 0.081. Hence present value is 10,000/(1.08110) =4589.26. (b) Expected profit is a4589.26(0.3× 1.0710 + 0.5× 1.0810 + 0.2× 1.110)− 10,000 = 42.94

29. Use(1 + i(4)/4

)4= 1 + i. For part (a) we have

(1 + i(4)/4

)4= 1.00512. Hence i(4) = 4

[1.0053 − 1

]= 0.0603 or

6.03%. For part (b) we have(1 + i(4)/4

)4= (1 + 2 × 0.06)1/2 = 1.121/2. Hence i(4) = 4

[1.121/8 − 1

]= 0.5707 or

5.707%.

30. First investor: £96 grows to £97.50 in 45 days. Hence 1 + i1 = (97.50/96)365/45.Second investor: £97.50 grows to £100 in 45 days. Hence 1 + i2 = (100/97.50)365/45.As 100/97.50 = 1.0256 > 97.50/96 = 1.0156, the second investor receives a higher rate of interest.

31. Capital gain is 31p. Hence CGT is 9.3p. Hence 769 grows to (800 − 9.3) + 4 = 794.7 in half a year. Hence1 + i = (794.7/769)2 giving i = 0.067957 or 6.80%.

Appendix 2 Sep 27, 2016(9:51) Answers 1.5 Page 147

32. (a)(i) Suppose n days. Then 4000 = 3600(1 + 0.06n

365

). Hence n = 365/(9 × 0.06) = 18250/27 = 675.9. Hence

676 days are needed. (ii) Now 1.0154t = 10/9. Hence 4t = ln(10/9)/ ln(1.1015). Or t = 0.25 ln(10/9)/ ln(1.1015)years, or 91.25 ln(10/9)/ ln(1.1015) = 645.7 days. So 646 days are needed. (iii) Now 1.00512t = 10/9. Ort = (1/12)× ln(10/9)/ ln(1.0005) years or (365/12)× ln(10/9)/ ln(1.0005) = 642.54 days. So 643 days are needed.(b) For time periods longer than 1 year, compound interest pays more because the investor receives the benefit ofinterest on interest.

33. (a) We need d with 3,000(1 + 0.04× d/365

)= 3,800. Hence d = 7,300/3 = 2,433.33 or 2,434 days. (b) We

need t with 3,000 × 1.04t = 3,800. Hence t = ln(19/15)/ ln(1.04). The number of days is 365t = 2199.9066 or2,200 days.

34. If i denotes the effective interest rate per annum, then 1 + i = 1.032. Let ν = 1/(1 + i) = 1/1.032; let ν1 = ν1/4 =1/1.031/2; and finally, let ν2 = ν1/12 = 1/1.031/6. Value of payments in first 4 years at time 0 is 120(1+ν +ν2 +ν3) =120(1− ν4)/(1− ν). Hence value at time t = 12 is 120(1− ν4)× 1.0324/(1− ν). Value of payments in next 4 yearsat t = 12 is 30(1−ν16

1 )×1.0316/(1−ν1). Value of payments in final 4 years at t = 12 is 10(1−ν482 )×1.038/(1−ν2).

Hence answer is 894.8930 + 691.0424 + 542.8388 = 2128.77.

Answers to Exercises: Chapter 1 Section 5 on page 14 (exs1-2.tex)

1. The discount factor is 1/(1 + 0.08× 91/365) = 0.980.

2. The discount factor is 1/(1 + 0.08)3 = 1/1.083 = 0.794

3. Now i = 25/300 and 1 + i = 13/12. Hence c0(1 + i)2 = 3380 giving c0 = 3380× 12/13× 12/13 = 2880.

4. (a) 720(1 + 1/6)2 = 980 (b) Using (1 + i)(1− d) = 1 gives d = 1/7 or 14.29%.

5. c1 = c0/(1− d) = 275/(1− 1/3) = 275× 3/2 = 412.5

6. By using the relation (1 + i)(1− d) = 1; i increases implies d increases and hence ν = 1− d decreases.

7. Let i1 = ki. Then d1 = i1/(1 + i1) and so d1/d = k(1 + i)/(1 + ki) < k.

8. (a) As i− d = i2/(1 + i), it follows that i− d > 0.

9. (a) Using(1− d(m)/m

)m= 1 − d with m = 1/2 and d(m) = 0.025 gives d = 1 −

√0.95 = 0.02532 or 2.53% p.a.

(b)(i) Now(1− d(4)/4

)4= 1 − d = 1/(1 + i) = 1/1.045. Hence d(4) = 0.0437756 or nominal 4.38% p.a. payable

quarterly. (ii) Using(1− d(12)/12

)12= 1 − d = 1/(1 + i) = 1/1.045 gives d(12) = 0.043936 or nominal

4.39% p.a. payable monthly. (iii) For this part, m = 1/3. Using(1− d(1/3)/(1/3)

)1/3= 1 − d = 1/(1 + i) =

1/1.045 gives 1− 3d(1/3) = 1/1.0453 and hence d(1/3) = 0.041234 or nominal 4.12% p.a. payable every 3 years.

10. (1− dep)1/p = 1− d = 1/(1 + i) = 1/(1 + ieq)1/q .

11. (a) (i) (1.1)k/12−1 (ii) (1.1)p−1. (b) The effective discount rate is d = i/(i+1) = 1/11 p.a. and 1−d = 10/11.Then use (1− dep)1/p = 1− d. (i) Now p = k/12. Hence dep = 1− (10/11)k/12. (ii) dep = 1− (10/11)p

12. Use(1− d(m)/m

)m= 1 − d. Hence 1 − d = (1 − 0.02)4 = 0.984. Hence

(1− d(2)/2

)2= 0.984. Hence d(2) =

2(1− 0.982) = 0.0792 or nominal 7.92% payable every 6 months.

13. (a) The accumulated value after k years is c0(1 + i)k. Hence we want (1 + i)k = 2 or k = ln 2/ ln(1 + i).(b) The accumulated value after k years is c0e

ik. Hence we want eik = 2 or k = ln 2/i.(c) Annual compounding: 69.66, 14.21, 10.24, and 7.27. Continuous compounding: 69.31, 13.86, 9.90, and 6.93.(d) Now na = ln 2/ ln(1 + i) and nb = ln 2/i. Also i− i2/2 < ln(1 + i) < i for i > 0.

14. (a) Now c0(1 + i2) = c0(1 + i1)2. (i) Hence i2 = 2i1 + i21 and so i2 > 2i1. (ii) Now c0 = (1− d2)c2 = (1− d1)2c2.Hence d2 = 2d1 − d2

1 and so d2 < 2d1.(b) Now ct = c0(1 + it) = c0(1 + i1)t. Hence (1 + i1) = (1 + it)1/t = (1 + is)1/s. Hence 1 + it = (1 + is)t/s. Usingt = 2s gives 1 + it = (1 + is)2 and hence it > 2is. Similarly, (1 − ds) = (1 − d1)s and (1 − dt) = (1 − d1)t imply1− dt = (1− ds)t/s. Using t = 2s gives dt < 2ds.

15. (a) We are given d1 = 0.01 for 1 month. Hence c0 = (1 − d1)c1/12 = (1 − d1)2c2/12 = (1 − d2)c2/12. Henced2 = 1− (1− d1)2 = 1− 0.992 = 0.0199. (ii) Similarly dk = 1− (1− d1)k = 1− 0.99k(b) 0.983 × 5000 = 4705.96

16. (a) Now(1 + i(m)/m

)m= 1 + i = eδ . Hence i(m) = m

(eδ/m − 1

). Using

(1 + i(m)/m

)m (1− d(m)/m

)m= 1

gives(1− d(m)/m

)m= e−δ and hence the result. (b) To show i > i(2) > i(3) > · · · see question 23 on page 9

(section 3 of chapter 1). To show d < d(2) < d(3) < · · ·, let g(m) = m(1−e−δ/m). Then g′(m) = 1−(1+δ/m)e−δ/m.But ex > 1 + x for x > 0 and hence g′(m) > 0 for m > 0. Hence result. (c) By exercise 22 on page 9,we have limn→∞ n(x1/n − 1) = lnx. Hence limm→∞ i(m) = δ. From

(1 + i(m)/m

) (1− d(m)/m

)= 1 we get

1/d(m) = 1/m + 1/i(m) and hence limm→∞ d(m) = limm→∞ i(m).


17. Using c0 = (1− nd)cn gives c0 =(1− 90× 0.06/365

)cn = 1798cn/1825. Also cn = (1 + i)90/365c0 where i is the

required answer. Hence i =(1825/1798

)365/90 − 1 = 0.06231 or effective 6.23% p.a.

18. Using c0 = (1 − nd)cn gives c0 =(1− 0.045× 28/365

)cn =

(1− 1.26

365

)cn = 36,374cn/36,500. Also cn =

(1+i)28/365c0 where i is the required answer. Hence i =(36,500/36,374

)365/28−1 = 0.04611 or effective 4.61% p.a.

19. Using c0 = (1 − nd)cn gives c0 = (1 − 90 × 0.05/365)cn = 721cn/730. Also cn = (1 + i)90/365c0 where i isthe effective rate p.a. and (1 + i(2)/2)2 = 1 + i where i(2) is the required answer. Hence i(2) = 2

[(1 + i)1/2 − 1

]=

2[(730/721)365/180 − 1

]= 0.05094891 or nominal 5.09% p.a. converted 6-monthly.

20. Using c0 = (1− nd)cn gives c0 = (1− 0.02)4c1 = 0.984c1.(a) Now c1 = (1 + i(2)/2)2 where i(2) is the required answer. Hence i(2) = 2(1/0.982 − 1) = 0.08246564 or nominal8.25% p.a. convertible 6-monthly. (b) Now c0 = (1 − d(12)/12)12c1 where d(12) is the required answer. Henced(12) = 12(1− 0.981/3) = 0.0805393 or nominal 8.05% p.a. convertible monthly.

21. (i)(a) Now c1 = 1.045c0 and we want d with c0 = (1−d)c1. Hence d = 0.045/1.045 = 0.04306 or 4.306%. (b) Wewant d(12) with c0 = (1−d(12)/12)12c1. Hence d(12) = 0.04394 or 4.394%. (c) We want i(4) with (1+i(4)/4)4 = 1.045.Hence i(4) = 0.04426 or 4.426%. (d) Now c5 = 1.0455c0 = 1.24618193765c0 and hence 24.618%.(ii) The discount rate d(12) is applied each month to a value which has already been discounted. Hence the rated(12)/12 must be higher than d/12 in order to obtain the same total amount of discounting.

22. (The question is not clear: you need to assume the treasury bill is also for 91 days.) Now if c0 is the price of thetreasury bill, then c0×1.0491/365 = 100. Hence if d is the annual simple rate of discount, then c0 = (1−91d/365)100.Hence

d =36591

(1− 1

1.0491/365

)0.0390296 or 3.90%.

23. (i) Suppose d days. Hence $96.50 grows to $98 in d days. Hence 96.5 × 1.04d/365 = 98 giving d/365 =ln(98/96.5)/ ln(1.04) and hence d = 143.54 or 144 days.(ii) The amount $98 grows to $100 in 38 days. Hence 98(1 + 38i/365) = 100 giving i = 0.196026 or 19.603%.(iii) We want i with 98(1 + i)38/365 = 100. Hence i = (100/98)365/38 − 1 = 0.21415980 or 21.416%.

24. (a) We need c0 with c0 × 1.0391/365 = 100. This gives c0 = 100/1.0391/365 = 99.26576. (b) We need d withc0 = (1− 91d/365)100. Hence d = (365/91)(1− 1/1.0391/365) = 0.029450153 or 2.945%.

25. (i) We have c0 = (1 − 0.055/12)12c1. (a) We need δ with c1 = eδc0. Hence eδ = (1 − 0.055/12)−12 givingδ = 0.0551264 or 5.513%. (b) We need i with 1 + i = (1 − 0.055/12)−12 and hence i = 0.0566741 or 5.667%.(c) We need i(12) with (1+ i(12)/12)12 = (1−0.055/12)−12 and hence i(12) = 12

[(1− 0.055/12)−1 − 1

]= 0.0552532

or 5.525%. (ii) In (b), the interest is just added at the end of the year; in (c), the interest is added every month andcompounded every month. Because the final sum is the same, the rate in (c) must be less than the rate in (b).(iii) We need d with 100 = (1 − d)8 × 159. Hence d = 0.0563187 or 5.632%. (iv) We need i with (1 + i)2 = 1.12.Hence i = 0.0583006 or 5.830%.


1. (a) 1000 exp(0.05/2) = 1025.31 (b) 1000× 1.050.5 = 1024.6951 or £1,024.70 to nearest penny.

2. 1000 exp(∫ 6

1 δ(s) ds)

= 1000 exp(∫ 6

1 (0.04 + 0.01s)ds)

= 1000 exp(0.2 + 0.175) = 1454.99

3. (a) δ(t) = −ν′(t)/ν(t) (b) b.

4. A(t) = eδt, ν(t) = e−δt and δ(t) = δ.

5. Differentiating δ(t) = 1t

∫ t0 δ(s) ds gives

dδ(t)dt

=1t2

(tδ(t)−

∫ t

0δ(s) ds

)for t > 0.

As δ(s) ↑ as s ↑, it follows that∫ t

0 δ(s) ds ≤ tδ(t). Hence result.

6. Recall that δ(t) = 1t

∫ t0 δ(s) ds. Hence δ(t) = −ln ν(t)/t. Suppose δ ↑ as t ↑. Then for all t > 0 and all α ∈ [0, 1] we

have δ(αt) ≤ δ(t) and so ln ν(αt) ≥ α ln ν(t). It follows that ν(αt) ≥(ν(t)

)α. The proof of the converse is similar.

7. Now ν(t) = exp(−∫ t

0 δ(s) ds)

= exp[−0.06(0.7t − 1)/ ln 0.7

]. The present value of £100 in 31/2 years’ time is

100× ν(3.5) = 88.696882 or £88.70.

8. (a) Use 1 + pip(t) = A(t + p)/A(t) and A(t) = exp(∫ t

0 δ(u) du)

.

(b) Now∫ t

0 au du = (at − 1)/ ln a. Hence pip(t) = exp(0.06× 0.7t(1− 0.7p)× ln 10

7 )− 1.


9. (a) Using cn = c0(1 + ni) gives 1,550 = 1,500(1 + 0.05n

365

). Hence n = 730/3 = 243.3 days. It takes 244 days.

(b) 1,550 = 1,500e0.05n/365. Hence n = 7,300× ln(31/30) = 239.37 days. It takes 240 days.

10. Now 210 = 200 exp(∫ 5

0 (a + bt2) dt)

= 200 exp(5a+125b/3). Also 230 = 200 exp(∫ 10

0 (a + bt2) dt)

= 200 exp(10a+1000b/3). So we have the 2 simultaneous equations: 5a + 125b/3 = ln(21/20) and 10a + 1000b/3 = ln(23/20).Hence 750b/3 = ln(23× 20/212) = ln(460/441); hence b = (3/750)× ln(460/441) = 0.00016872646 or 0.0001687.Similarly, 30a = ln(218/(207 × 23)) and so a = 0.008351979 or 0.008352.

11. (a)(i) Using c0(1 + 0.05/12

)120= 100 gives c0 = 60.716 or £60.72. (ii) For this case, c0 =

(1− 0.05/12

)120×100.Hence £60.59. (iii) Using c0e

0.5 = 100 gives c0 = 100e−0.5 = 60.6531 or £60.35. (b). Let i denote theequivalent effective annual interest rate. Hence (1 + i)91/365 = 100/98.91. Hence i = 0.04494 or 4.49%.

12. Now∫ t

0 δ(s) ds equals 0.04t + 0.005t2 if t < 8 and 0.64 + (t − 8)0.07 = 0.08 + 0.07t if t > 8. Hence A(t) equalse0.04t+0.005t2

if t < 8 and e0.08+0.07t if t > 8.(b) The present value is 100/A(10) = 100e−0.78 = 45.84.

13. The general formula is A(t2) = A(t1) exp(∫ t2

t1δ(s) ds

). So we want 500 exp

(∫ 102 δ(s) ds

)+ 800 exp

(∫ 109 δ(s) ds

).

Now∫ 10

9 δ(s) ds =∫ 10

9 (0.01s − 0.03)ds = 0.065 and∫ 10

2 δ(s) ds = 0.345. Hence we want 500e0.345 + 800e0.065 =1559.72. (b) We want i where 500(1 + i)8 + 800(1 + i) = 1559.72. Now £1 grows to £e0.345 = £1.412 in 8 yearswhich corresponds to 4.4%. So guess 4%: then 500(1 + i)8 + 800(1 + i) = 1516.28.Guess 5%: then 500(1 + i)8 + 800(1 + i) = 1578.73. So answer is 5% to nearest 1%. (It is 4.71%.)

14. (i) Now c0 = (1 − 0.02)4c1 = 0.984c1 and c1 = eδc0. Hence δ = −4 ln 0.98 = 0.0808108. (ii) Now c1 = (1 + i)c0 =(1+i)0.984c1. Hence i = 1/0.984−1 = 0.0841658. (iii) Now c0 = (1−d(12)/12)12c1. Hence d(12) = 12(1−0.981/3) =0.0805393.

15. Now∫ t

0δ(s) ds =

0.05t + 0.01t2 for 0 ≤ t ≤ 5;0.15t− 0.25 for t > 5

and hence ν(t) =e−0.05t−0.01t2

for 0 ≤ t ≤ 5;e0.25−0.15t for t > 5

(i) We want 1,000ν(12) = 1,000e−1.55 = 212.24797383 or £212.25.(ii) We want d with 212.25 = (1− d)121,000 and hence d = 0.12103392 or 12.1%.


1. (a) NPV(a) = 2575.25 and NPV(b) = 2509.90.(b) NPV(a) = 2372.73 and NPV(b) = 2509.09.(c) At time 0, borrow 2400 and take c0 = b0 = 2500. This transforms a = (100, 2500) intoc = (2500,−2400(1 + 0.01) + 2500 = 76).(d) At time 0, invest 2400 and take c0 = a0 = 100. This transforms b = (2500, 10) intoc = (100, 2400(1 + 0.1) + 10 = 2650).(e) Consider the two cases a0 ≥ b0 and a0 < b0 and use induction on n.If a0 ≥ b0 then set c0 = b0 and invest a0− b0. This converts a into the sequence (b0, (1 + i)(a0− b0) + a1, a2, . . . , an).Let a∗ = ((1 + i)(a0 − b0) + a1, a2, . . . , an) and b∗ = (b1, . . . , bn). Then

NPV(a∗)− NPV(b∗) = (1 + i)(a0 − b0) +n−1∑j=0

aj+1

(1 + i)j−n−1∑j=0

bj+1

(1 + i)j

= (1 + i) [NPV(a)− NPV(b)]≥ 0

Hence result by induction assumption. Similarly for a0 < b0.

2. The value at time 0 is:

NPV = A(0) +∫ t

0ρ(s)ν(s) ds where ν(s) = exp

[−∫ s

0δ(u) du

]Hence value at time t is:

NPV× exp[∫ t

0δ(u) du

]= A(0) exp

[∫ t

0δ(u) du

]+∫ t

0ρ(s) exp

[∫ t

s

δ(u) du]ds

Alternatively: NowA′(t) = δ(t)A(t)+ρ(t). As usual, let ν(t) = e−∫ t

0δ(s) ds denote the value at time 0 of the amount 1

at time t. Integrating the equation ν(t)A′(t) = ν(t)δ(t)A(t) + ν(t)ρ(t) gives

ν(t)A(t)− ν(0)A(0) =∫ t

0ν(s)ρ(s) ds

and hence


A(t) =A(0)ν(t)

+∫ t

0

ν(s)ν(t)

ρ(s) ds = A(0)e∫ t

0δ(s) ds +

∫ t

0e

∫ t

sδ(h) dh

ρ(s) ds

(b) A(t) = A(0)(1 + i)t +∫ t

0 (1 + i)t−sρ(s) ds

3. Using equation (2.9b) gives

NPV = 4∫ 1

0νu du + 6

∫ 2

1νu du + 8

∫ 3

2νu du

But∫νu du =

∫exp(u ln ν) du = νu/ ln ν.

Hence NPV =[4(ν − 1) + 6(ν2 − ν) + 8(ν3 − ν2)

]/ ln ν =

[8ν3 − 2ν2 − 2ν − 4

]/ ln ν. This leads to A(3) =

NPV/ν3 =[4× 1.043 + 2× 1.042 + 2.08− 8

]/ ln 1.04 = 18.935.

4. Now ρ = 100 and NPV =∫ 6

0 ρ(u)ν(u) du = 100∫ 6

0 ν(u) du where ν(t) = exp[−∫ t

0 δ(u)du]. Hence NPV =

100∫ 6

0 exp [−0.06t] dt = 100(1 − e−0.36)/0.06. Then use A(12) = NPV × exp[∫ 12

0 δ(u) du]. where

∫ 120 δ(u) du =

0.06× 6 +∫ 12

6 (0.05 + 0.0002s2) ds = 0.36 + 0.3 + 0.1008 = 0.7608. Hence A(12) = NPV× exp(0.7608) = 1078.28.

5. Solving 990(1 + i)− 65(1 + i)0.5 − 1076 = 0 gives (1 + i)0.5 = 1.075875 and so i(2) = 2[(1 + i)0.5 − 1

]= 0.15175 or

15.17%.

6. Using equation (2.9a) gives NPV =∫ 8

0 50ν(u) du. But for 0 ≤ t ≤ 8 we have ν(t) = exp(−∫ t

0 δ(u) du)

=

exp(−0.05t). Hence NPV = 50∫ 8

0 exp(−0.05u) du = 50(1− e−0.4)/0.05 = 1,000(1− e−0.4). Value at time t = 15 isA(15) = NPV/ν(15). Now

∫ 150 δ(u) du = 0.4+

∫ 158 (0.04+0.0004u2) du = 0.68+0.0004(153−83)/3 = 0.68+1.1452/3.

Hence A(15) = 953.2295 or £953.23.

7. Use equation (2.9a) with ρ(t) = 50 exp(0.01t) for 9 < t < 12. Hence NPV =∫ 12

9 50 exp(0.01u)ν(u) du whereν(u) = exp

(−∫ u

0 δ(s) ds). Now for u > 8 we have

∫ u0 δ(s) ds = 0.16+0.08+0.48−0.01×24+0.06(u−8) = 0.06u.

Hence ν(u) = e−0.06u for u > 8. Hence NPV = 50∫ 12

9 e−0.05u du = 1,000(e−0.45 − e−0.6) = 1,000e−0.6(e0.15 − 1) =88.8165.

8. Let ν = 1/1.09 and use units of £1,000. Then NPV = −60 − 25ν2/3 + 50ν22 +∑4j=0

∫ 4j+64j+2 (4j + 5)νt dt = −60 −

25ν2/3 + 50ν22 +∑4j=0(4j + 5)ν4j+2

∫0 ν

t dt = −60− 25ν2/3 + 50ν22 + ν2(5 + 9ν4 + 13ν8 + 17ν12 + 21ν16)x where

x =∫ 4

0 νt dt = (ν4−1)/ log(ν). Hence NPV = −60−25ν2/3+50ν22+ν2(−5−4ν4−4ν8−4ν12−4ν16+21ν20)/ ln(ν) =

7.154485046 or £7,154.49.

9. (a) Use equation (6.1c). A(10) = 500 exp(∫ 10

0 δ(t) dt)

where∫ 10

0 δ(t) dt = 0.56− 0.005t2

2

∣∣∣80

+ 0.12 = 0.68− 0.16 =

0.52. Hence A(10) = 500e0.52 = 841.014 or £841.01. (b) Use equation (2.9a) with ρ(u) = 200e0.1u for 10 < u ≤18. Also, for u > 8 we have

∫ u0 δ(s) ds = 0.4 + 0.06(u− 8) = 0.06u− 0.08 and hence ν(u) = exp (−0.06u + 0.08).

Hence NPV =∫ 18

10 ρ(u)ν(u) du = 200∫ 18

10 exp(0.1u) exp(−0.06u + 0.08) du = 200e0.08∫ 18

10 e−0.04u du = 200e0.08 ×[

e0.72 − e0.4]/0.04 = 5,000

[e0.8 − e0.48

]= 3,047.33.

10. Let i denote the effective interest rate per annum; hence 1 + i = 1.024 and ν = 1/(1 + i) = 1/1.024. Let x = ν1/12.Then

NPV = 5,000

[∫ 1

0νu du +

ν

12(1 + x + x2 + · · · + x11) +

ν2

2(1 + ν1/2)

]

= 5,000

[∫ 1

0eu ln ν du +

ν

121− x12

1− x+ν2

2(1 + ν1/2)

]= 5,000

[eu ln ν

ln ν

∣∣∣∣u=1

u=0+ν

121− ν1− x

+ν2

2(1 + ν1/2)

]

= 5,000[ν − 1ln ν

+ν

121− ν1− x

+ν2

2(1 + ν1/2)

]= 13,347.39

11. (i) Use equation (6.2a):∫ 12

0 δ(s) ds = 0.2 + 0.04t2∣∣t=10t=5 + (0.0025t2 + 0.0001t3)

∣∣t=12t=10 = 0.2 + 0.3 + 0.005 × 22 +

0.0001 × 728 = 0.6828. Hence NPV = ν(12) = e−0.6828 = 0.5052. (ii) We want i with (1 + i)12 = e0.6828, ori = exp(0.6828/12)− 1 = 0.0585499501 or 5.85%. (iii) Use equation (2.9a). For 2 < u < 5, ρ(u) = e−0.05u and∫ u

0 δ(s) ds = 0.04u. Hence

NPV =∫ 5

2e−0.09u du = −e

−0.09u

0.09

∣∣∣∣52

=e−0.18 − e−0.45

0.09= 2.196


12. (i) Use equation (6.1c). We have 150 = 100 exp(∫ 5

0 δ(s) ds)

= 100 exp( 25a

2 + 125b3

)or 25a

2 + 125b3 = ln

( 32

). Similarly

and 50a+ 1,000b3 = ln

( 2310

). Hence 500b

3 = ln( 23

10

)−4 ln

( 32

)= ln

( 2310

)− ln

( 8116

)= ln

( 184405

). Hence b = 3

500 × ln( 184

405

)=

−0.004734. Also 50a = 8 ln( 3

2

)− ln

( 2310

)= ln

( 328052944

)and hence a = 0.048216.

(ii) We want δ with 100e10δ = 230 or δ = (1/10)× ln(23/10) = 0.08329.(iii) Use equation (2.9a) with ρ(u) = 20e0.05u for 0 < u < 10 and ν(u) = e−δu where δ is given in part (ii). Hence

NPV =∫ 10

0ρ(u)e−δu du = 20

∫ 10

0e(0.05−δ)u du = 20

1− e0.5−10δ

δ − 0.05=

20023

23− 10e0.5

ln(23/10)− 0.5= 170.1153

where we have used e−10δ = 10/23 and 10δ = ln(23/10) from part (ii).

13. (i) Now ∫ 10

0δ(s) ds = 0.2 + 0.01

(t3

3− t2

2

)∣∣∣∣t=10

t=5= 0.2 +

15.256

Hence ν(10) = exp(−0.2− 15.25/6

)= 0.06446282 or 0.0645.

(ii) Now 1 + i = ν(9)/ν(10) = A(9, 10) = exp[∫ 10

9 0.01(s2 − s) ds]

= exp[0.01× 485/6

]= 2.24416 and hence

i = 1.24416 or 124.16%.(iii) (a) ν(t) = exp

(−∫ t

0 δ(s) ds)

= exp (−0.04t) (b) NPV(ρ) =∫ 5

0 ρ(u)ν(u) du =∫ 5

0 1 du = 5.

14. (i) If t ≤ 8,∫ t

0 δ(s) ds =∫ t

0 (0.03 + 0.01s) ds = 0.03t + 0.005t2. At t = 8 this has value 0.56. Hence for t > 8 wehave

∫ t0 δ(s) ds = 0.56 + (t − 8)0.05 = 0.05t + 0.16. Hence ν(t) = exp(−0.03t − 0.005t2) if 0 ≤ t ≤ 8 and ν(t) =

exp(−0.16 − 0.05t) if t > 8. (ii) (a) We want 500ν(15) = 500 exp(−0.16 − 0.75) = 500 exp(−0.91) = 201.26.(b) let d(4) be the required quantity. Now 201.26 grows to 500 in 15 years. Hence 500

[1− d(4)/4

]60= 201.26 and

d(4) = 0.0602096 or 6.02%. (iii) We want∫ 14

10 ρ(t)δ(t) dt =∫ 14

10 10e−0.02te−0.16−0.05t dt = 10e−0.16∫ 14

10 e−0.07t dt =

10[e−0.86 − e−1.14

]/0.07 = 14.763.

15. (i) If t ≤ 10,∫ t

0 δ(s) ds =∫ t

0 (0.04 + 0.01s) ds = 0.04t + 0.005t2. At t = 10 this has value 0.9. Hence for t > 10we have

∫ t0 δ(s) ds = 0.9 + (t − 10)0.05 = 0.05t + 0.4. Hence ν(t) = exp(−0.04t − 0.005t2) if 0 ≤ t ≤ 10 and

ν(t) = exp(−0.4 − 0.05t) if t > 10. (ii)(a) We want 1000ν(15) = 1000 exp(−0.4 − 0.75) = 1000 exp(−1.15) =316.636769 or £316.64. (b) 1000(1− d)15 = 316.64. Hence d = 1− exp(−1.15/15) = 0.07380 or 7.38%.(iii) We want

∫ 1510 ρ(t)δ(t) dt =

∫ 1510 20e−0.01te−0.4−0.05t dt = 20e−0.4

∫ 1510 e−0.06t dt = 20e−0.4

[e−0.6 − e−0.9

]/0.06 =

1000(e−1 − e−1.3)/3 = 31.78255.

16. (i) For t > 6 we have ∫ t

0δ(s) ds = 0.6− 0.09 + 0.07(t− 6) = 0.09 + 0.07t

Hence ν(7) = exp(−0.58) and ν(10) = exp(−0.79). Answer is 100/ν(10) + 50ν(7)/ν(10) = 100 exp(0.79) +50 exp(0.21) = 282.0235456 or £282.02.(ii) NPV =

∫ 1512 ρ(u)ν(u) du = 50

∫ 1512 exp(−0.09− 0.02t) dt = 2500(e−0.33 − e−0.39) = 104.667147 or £104.67.

17. (i) Now the value at time t = 0 of £1,000 at time t = 10 is 1,000ν(10) and this has value 1,000ν(10)/ν(5) at timet = 5. Recall ν(t) = exp

(−∫ t

0 δ(s) ds)

. Hence we want 1,000 exp(−∫ 10

5 δ(s) ds)

.

Now∫ 10

5 δ(s) ds =∫ 7

5 (0.1− 0.01s) ds +∫ 10

7 (0.01s− 0.04) ds = 0.2− 0.12 + 0.51/2− 0.12 = 0.215. Hence answeris 1,000 exp(−0.215) = 806.541440 or £806.54.(ii) The equivalent effective rate of interest per annum is given by 806.54(1 + i)5 = 1,000. Hence we want i(12) with806.54(1 + i(12)/12)60 = 1,000. Hence i(12) = 0.043077 or 4.3%.(iii) Value at time t = 0 is

∫ρ(s)δ(s) ds =

∫ 40 100e0.02se−0.06s ds = 100

∫ 40 e−0.04s ds = 2500(1 − e−0.16). We want

the value at time t = 12 which is 2500(1− e−0.16)/ν(12). Now∫ 12

0 δ(s) ds = 0.24 + 0.3− 0.165 + 0.475− 0.2 = 0.65and hence ν(12) = exp(−0.65). Hence required answer is 2500(1− e−0.16)× exp(0.65) = 708.061523 or £708.06.

18. (i) Now∫ 10

0 δ(s) ds = 0.2 + 0.003 t3

3

∣∣∣50

+ 0.03 + 0.03 t2

2

∣∣∣85

+ 0.04 = 0.98. Hence NPV = 1,000ν(10) = 1,000 exp−0.98 =

375.311099 or £375.31. (ii) Now 1,000(1− d(4)/4)40 = 375.31. Hence d(4) = 0.09680953 or 9.68%.(iii) Now NPV =

∫ 1810 ρ(u)ν(u) du =

∫ 1810 100e0.01uν(u) du. For u > 10 we have

∫ u0 δ(u) du = 0.02u + 0.78 and

so ν(u) = e−0.02u−0.78. Hence NPV = 100e−0.78∫ 18

10 e−0.01u du = 10,000e−0.78(e−0.1 − e−0.18) = 318.900257 or

£318.90.


19. (i) If 0 ≤ t ≤ 9 then∫ t

0 δ(s) ds = 0.03t + 0.005t2. Hence∫ 9

0 δ(s) ds = 0.27 + 0.405 = 0.675. Hence for t > 9 wehave

∫ t0 δ(s) ds = 0.675 + 0.06(t− 9) = 0.135 + 0.06t.

Hence ν(t) = e−0.03t−0.005t2for 0 ≤ t ≤ 9 and ν(t) = e−0.135−0.06t for t > 9.

(ii)(a) We need 5,000ν(15) = 5,000e−0.135−0.9 = 5,000e−1.035 = 1776.131905 or £1,776.13.(b) We need δ with 1,776.13e15δ = 5,000. Hence δ = ln(5,000/1,776.13)/15 = 0.06900007148965 or 0.069.(iii) Now NPV =

∫ 1511 ρ(s)ν(s) ds =

∫ 1511 100e−0.02se−0.135−0.06s ds = 100

∫ 1511 e−0.135−0.08s ds = 100e−0.135(e−0.88 −

e−1.2)/0.08 = 124.055318 or 124.055.

20. Now∫ t

0 δ(s) ds = 0.03t + 0.0025t2 for t < 3 and 0.0975 + 0.005t for t > 3. Hence NPV = 5,000∫ 6

3 ν(u) du =5,000

∫ 63 e−0.0975−0.005u du = 5,000e−0.0975

[e−0.015 − e−0.03

]/0.005 = 1,000,000

(e−0.1125 − e−0.1275

). This has

value 13303.9312743 or £13,303.93.(ii) We need i with 1 + i = eδ where 13,303.93 = 5,000

∫ 63 e−δu du. Hence 13,303.93 = 5,000

[e−3δ − e−6δ

]/δ.

Definef (i) = 13,303.93− 5,000[e−3δ − e−6δ

]/δ where δ = ln(1 + i). Using the approximation ex ≈ 1 + x + x2/2

shows[e−3δ − e−6δ

]/δ is approximately 3−14δ. This leads to i ≈ 0.025. Now f (0.025) = −121.65 and f (0.03) =

167.84. Linear interpolation gives i ≈ 0.0271 or 2.7%. (Actually 2.70844%.)(iii) 300A(50) = 300 exp(0.0975 + 0.005× 50) = 300e0.3475 = 424.657293 or £424.66.


1. The following function does more than the question requires. It also returns the interest rate and NPV just before thesign changes (if one exists).

"irr" <- function(vec.cashflow, r)n <- length(vec.cashflow)numr <- length(r)tmp <- outer(1/(1 + r), 0:(n - 1), FUN = "^")NPV <- c(tmp %*% vec.cashflow)signs <- sign(NPV)signchanges <- (1:numr)[(signs[ - numr] * signs[-1]) <= 0]if(length(signchanges) == 0)noroot <- T

else noroot <- Ffirstchange <- signchanges[1]

results <- cbind(interest = r, NPV = NPV)if(noroot) list(results)

else list(table = results, interest = r[firstchange], NPV = NPV[firstchange])

2. Applying the bond yield to the cash flow of project A gives: −50 + 8/1.07 + 8/1.072 + · · · + 8/1.078 = −2.23. Thissuggests that project A is not profitable.

3. Tke IRR is the solution for r of the equationm∑j=0

cj(1 + i)tj

=n∑j=0

bj

(1 + i)t′j

Define the function g : (−1,∞)→ R by

g(r) =n∑j=0

bj

(1 + r)t′j−tm−

m∑j=0

cj(1 + r)tm−tj

Then g is continuous with limr→−1 g(r) = ∞, limr→∞ g(r) = −∞ and g(r) ↓↓ as r ↑↑ for r ∈ (−1,∞). Hencethere exists a unque r∗ ∈ (−1,∞) with g(r∗) = 0 and then

m∑j=0

cj(1 + r∗)tj

=n∑j=0

bj

(1 + r∗)t′j


4. Use units of £1,000. We have:

Time 0 8 9 10 11Cash flow for project A (£ million) -1,000 1,700 0 0 0Cash flow for project B (£ million) -1,000 1,000 321 229 245

(i) Let i′ denote the required effective interest rate per annum and let ν = 1/(1+i′). Then 1,700ν8 = 1,000ν8+321ν9+229ν10+245ν11 or 700(1+i)3−321(1+i)2−229(1+i)−245 = 0. So let f (i) = 700(1+i)3−321(1+i)2−229(1+i)−245.Now the return on project A is 7%. So try f (0.07) = −0.0128 and f (0.071) = 1.4774767. Interpolation givesi′ = 0.07 + 0.001× 0.0128/(0.0128 + 1.4774767) = 0.7001 or 7.00%.(ii) The cash is received earlier with project A than with project B. Hence an interest rate higher than i′ favoursproject A; an interest rate lower than i′ favours project B.In particular, if i = 0.06 then NPVA = −1,000 + 1,700/1.068 = 66.6010 and NPVB = −1,000 + 1,000/1.068 +321/1.069 + 229/1.0610 + 245/1.0611 = 74.3471.

5. The annual effective interest rate is i = 1.032 − 1 = 0.609; so let ν = 1/1.0609. Here are the discrete cash flows:

Time 1 Jan 2001 1 August 2001 31 Dec 2011Time in years 0 7/12 11Cash flow (£ million) -2 -1.5 3

We also have the continuous cash flow of 0.3 for t ∈ (1, 2), 0.4 for t ∈ (2, 3), . . . , 1.2 for t ∈ (10, 11). Hence

NPV(c, ρ) = −2− 1.5ν7/12 + 3ν11 + 0.3∫ 2

1νt dt + · · · + 1.2

∫ 11

10νt dt

= −2− 1.5ν7/12 + 3ν11 +12∑k=3

0.1k∫ k−1

t=k−2νt dt

= −2− 1.5ν7/12 + 3ν11 +12∑k=3

0.1k[νk−1 − νk−2

ln ν

]= −2− 1.5ν7/12 + 3ν11 +

0.1ln ν

[−3ν − ν2 − ν3 − · · · − ν10 + 12ν11]

= −2− 1.5ν7/12 + 3ν11 +0.1

(1− ν) ln ν[−3ν + 2ν2 + 13ν11 − 12ν12]

= −1.883476 + 4.9922 = 3.10875427438 or £3,108,754.

6. Let ν = 1/1.09. The NPV of outgoings (in £ million) is:

NPVO = 9 + 12∫ 2

1νu du = 9 +

12νu

ln ν

∣∣∣∣2u=1

= 9 +12(ν2 − ν)

ln ν= 9− 12× 900

1092 ln ν= 9 +

10,80011,881 ln 1.09

Hence NPVO = 19.5481398596. Now let x =√v = 1/1.091/2. Incomings (in £ million):

Time 1/1/06 1/7/06 1/1/07 1/7/07 1/1/08 1/7/08 1/1/09 · · · 1/1/21Time in years 0 1/2 1 1.5 2 2.5 3 · · · 15Cash flow (£ million) 0 0 0 0 0 2.5 2.5 · · · 2.5

After k payments, NPVI = 2.5[x5 + x6 + · · · + xk+4] = 2.5x5(1− xk)/(1− x). Setting NPVI = NPVO gives

k∗ = ln[

1− NPVO(1− x)2.5x5

]/ln(x) = 12.2 or the 13th payment.

This gives a DPP of 8.5 years after 1/1/06; that means the payment on 1 July 2014.(ii) Incoming payments are received later than the outgoings. Hence, increasing the interest rate will reduce NPVImore than NPVO. Hence the DPP will increase.

7. (i) Let ν = 1/1.1. Then NPVO = 10∫ 2

0 νt dt = 10 νt

ln ν

∣∣∣20

= 10(v2−1)ln ν .

Now NPVI = 8ν3 + 7.5ν4 + 7ν5 + 6.5ν6 + · · · + 4ν11 + 3.5ν12 + 3ν13 + 2ν14 + ν15 = x + 2ν14 + ν15 where 2(1− ν)x =16ν3 − (ν4 + · · · + ν13)− 6ν14 = ν3(16− 17ν − 5ν11 + 6ν12)/(1− ν). Hence −NPVO + NPVI = 14.5916944289 or£14,591,694.(ii) Incomings are received later than outgoings. Hence increasing i will decrease NPVI more than NPVO. Hencethe IRR is more than 10%. (It is 23.7%.)

8. Let α = 1.05 and ν = 1/1.09. The cash flows are as follows:Time 1/1/00 1/7/00 1/1/01 1/1/02 31/12/02 1/1/03 31/12/03Cash flow −18 −10 −5 −45 60.5 −45α 60.5αTime 1/1/04 31/12/04 1/1/05 31/12/05 1/1/06 31/12/06 1/1/07 31/12/07Cash flow −45α2 60.5α2 −45α3 60.5α3 −45α4 60.5α4 −45α5 60.5α5


After k years production, we have NPVO = 18 + 10ν1/2 + 5ν + 45ν2[1 + αν + · · · + (αν)k−1] = 18 + 10ν1/2 + 5ν +45ν2[1− (αν)k]/[1− αν]. Also NPVI = 60.5ν3[1 + αν + · · · + (αν)k−1] = 60.5ν3[1− (αν)k]/[1− αν]. EquatingNPVO = NPVI gives k = ln

[1− (18 + 10ν1/2 + 5ν)(1− αν)/(60.5ν3 − 45ν2)

]/ln(αν) = 3.832792 or 4 years

production. This gives a date of 31/12/05 which is 6 years after the start of the project.(ii) Incomings are received later than outgoings. Hence decreasing i will increase NPVI more than NPVO. Hencethe DPP will decrease.

9. Use units of £1,000. Let α = 1.06 and ν = 1/1.07.(i) Then NPVO = 180(1 + ν + ν2). We also have the continuous cash flow of 25αt for t ∈ (0, 25). Hence

NPVI =∫ 25

025αtνt dt = 25

(αν)25 − 1ln(αν)

Hence NPV = NPVI − NPVO = 51.61779346 or £51,617.80.(ii) We want twith NPVO =

∫ t0 25(αν)u du = 25[(αν)t−1]/ ln(αν). Hence t = ln

[NPVO ln(αν)/25 + 1

]/ ln(αν) =

ln[7.2(1 + ν + ν2) ln(αν) + 1

]/ ln(αν) = 22.4204905995 or 22.42 years.

(iii) We want to find α1 with

NPVO = NPVI and so 180(1 + ν + ν2) = 25(α1ν)25 − 1

ln(α1ν)Hence ν25α25

1 − 7.2(1 + ν + ν2) ln(να1) − 1 = 0. If α1 = 1.05 then LHS = 0.00541105; if α1 = 1.06 then LHS =−0.01938711. Linear interpolation gives α1 = 1.05 + 0.01f (1.05)/(f (1.05) − f (1.06)) = 1.052182 or 5.2%. Thenf (1.052) = −0.00266205. Linear interpolation gives α1 = 1.05 + 0.002f (1.05)/(f (1.05)− f (1.052)) = 1.05134051or 5.1%. In fact it is 5.13%.)

10. (i) The existence of the DPP just shows the project is profitable. It does not show how profitable the project is.

(ii) For project A we have

NPV = −1 +3.5

1.0410 = 1.3644746 or £1,364,475.

For project B, let ν = 1/1.04. Then we have

NPV = −1 +9∑j=0

∫ j+1

j

(0.08 + 0.01j)νt dt = −1 +9∑j=0

(0.08 + 0.01j)∫ j+1

j

νtdt

= −1 +9∑j=0

(0.08 + 0.01j)νjν − 1log ν

= 0.0073097664

or £7,309.77.(iii) DPP of project A is 10 (no payments are received before 10). DPP of project B is less than 10.(iv) Project A is preferable because it has the higher NPV (even though the DPP is larger).

11. Let α = 1.04. Then we have (in £million):Date 1/1/14 1/7/14 1/1/15 1/1/16 1/1/17 1/1/18 1/1/19 1/1/20 1/1/21 31/12/21Cash flow -19 -9 -5 -57 −57α −57α2 −57α3 −57α4 −57α5

75.6 75.6α 75.6α2 75.6α3 75.6α4 75.6α5

Let ν = 1/1.09. Assume production runs for k years where 1 ≤ k ≤ 6 or 31/12/16 to 31/12/21. Then

NPV = −19− 9ν1/2 − 5ν − 57k∑j=1

αj−1νj+1 + 75.6k∑j=1

αj−1νj+2

= −19− 9ν1/2 − 5ν + (75.6ν3 − 57ν2)k∑j=1

(αν)j−1

= −19− 9ν1/2 − 5ν + (75.6ν3 − 57ν2)1− (αν)k

1− αν

giving

(αν)k ≤ 1− (19 + 9ν1/2 + 5ν)(1− αν)75.6ν3 − 57ν2 = 0.857959

which gives k = 4 years of production or 31/12/19.(ii) If we decrease the interest rate from 9%, then ν increases. The DPP will be shorter. In the extreme, if the interestrate is 0% and so ν = 1, it would take just 2 years to recover the initial costs and so the DPP would be 2 years ofproduction. At the other extreme, if the interest rate was very high, it would never be possible to recover the costs ofborrowing the initial amounts which are invested in the project.


12. (i) (a) The DPP is the smallest time such that the accumulated value of the project becomes positive. Equivalently, itis the smallest time t such that the NPV of payments up to time t is positive. (b) The payback period is the timewhen the net cash flow (incomings less outgoings) is positive. The DPP and payback period are measures of the timeit takes for a project to become profitable.(ii) The payback period ignores discounting—and hence ignores the effect of interest. It is therefore an inferiormeasure and should not be used. Drawbacks of the DPP:• It shows when the project becomes profitable using a forecast interest rate, but it does not show how large (orsmall) the profit is.• Suppose the DPP of project A is less than the DPP of project B; it is possible that the NPV of project A is lessthan the NPV of project B—this can occur if there are is a large late cash inflow in project B.(iii) Take time 0 to be 1 January 2004. Let ν = 1/1.1. Then

NPV of making bid = 0.2

NPV of building costs =∫ 7

2νu du NPV of running costs =

∫ 7.25

7νu du

Total NPV of costs = 0.2 +∫ 7.25

2νu du = 0.2 +

ν7.25 − ν2

ln ν= 3.613811

NPV of TV rights = 0.3∫ 7.25

7νu du = 0.3

ν7.25 − ν7

ln ν= 0.0380320

Now for the other revenue:Year 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015Time in years 0.5 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5 10.5 11.5Cash flow 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.8 0.6 0.4 0.2

This has NPV:

NPV =ν0.5

10[1 + 2ν + 3ν2 + · · · + 12ν11 − ν8 − 4ν9 − 7ν10 − 10ν11]

=ν0.5

10[1 + 2ν + 3ν2 + · · · + 12ν11 − ν8(1 + 4ν + 7ν2 + 10ν3)

]=ν0.5

10

[1− ν12

(1− ν)2 −12ν12

1− ν− ν8

(1− 10ν4

1− ν+ 3ν

1− ν3

(1− ν)2

)]=ν0.5

10

[1 + 2ν12 − 3ν9

(1− ν)2 − 2ν12 + ν8

1− ν

]= 3.052965

Hence we need C with Cν11 = 3.613811− 3.052965− 0.038032 leading to C = 1.49165 or £149.165 million.

13. Use units of £1,000. Let ν = 1/1.08. For the discrete payments we have (in thousands) NPVO = 105 + 105ν1/2 +105ν + 200ν15. Let N1 denote the net present value of the continuous income for years 2, 3 and 4. Then

N1 = 20∫ 2

1νt dt + 23

∫ 3

2νt dt + 26

∫ 4

3νt dt

=20(ν2 − ν)

ln ν+

23(ν3 − ν2)ln ν

+26(ν4 − ν3)

ln ν=ν − 1ln ν

[20ν + 23ν2 + 26ν3]

For k ∈ 5, 6, . . . , 30, let N(k) denote the net present value of the continuous income for years 5, . . . , k. Letα = 1.03. Then

N(k) =k∑j=5

29αj−5∫ j

j−1νt dt =

29(ν − 1)ln ν

k∑j=5

αj−5νj−1 =ν − 1ln ν

29ν4 1− (αν)k−4

1− ανLet NPV(k) = −NPVO + N1 + N(k). (i) We need NPV(30) = −NPVO + N1 + N(30) = 4.30097626. Answeris £4,300.98 (ii) Now NPV(15) = −NPVO + NPV1 + NPV(15) = −129.57469246 and even if we remove the£200,000 payment (which has net present value equal to £63,048.34) from NPVO, this quantity is still negative.Also NPV(2) < NPV(3) < · · · < NPV(15) and so these are all negative. (iii) NPV(29) = −1.97144191 < 0.Hence the DPP occurs in the final year.


1. We use equation (7.4a) which is:

(1 + i)t =ft1−0

f0

ft2−0

ft1

· · · ftn−0

ftn−1

ftftn

Now t = 3, t1 = 0.5, etc. Hence

1.113 =460400× 550

510× 500

550× 600

540× 650

600× X

710=

460400× 500

510× 650

540× X

710and hence X = 715.50062864 or £715,500,629.


2. Using equation (7.4a) gives

(1 + i)3 =80120× 200

110× 200

210=

800693

leading to i = 0.049024 or 4.90%.

3. The TWRR, i is given by

(1 + i)3 =212180

261237

230261

273248

295273

309311

=212180

230237

295248

309311

= 1.35086

Hence i = 0.10544 or 10.54%.4. Use units of £1,000.

Time, t 1 Jan 2001 1 Nov 2001 1 May 2002 31 Dec 2002Fund value at time t− 0 137 173 205Net cash flow at time t 20 48Fund value at time t + 0 120 157 221

Hence the TWRR, i, is given by

(1 + i)2 =137120× 173

157× 205

221=

4,858,7054,163,640

Hence i = 0.080248519 or 8.02%.(ii) The disadvantage of the MWRR is that it is sensitive to the amounts and timings of cash flows—and the invest-ment manager has no control over these. The disadvantage of the TWRR is that it requires a lot more information—itrequires knowledge of all cash flows and the the value of the fund at the times of all cash flows.

5. Use units of £1,000,000. Let x denote the required amount. ThenTime, t 31 Dec 2001 31 Dec 2003 31 Dec 2004Fund value at time t− 0 x 4.2Net cash flow at time t 1.44Fund value at time t + 0 2.2 x + 1.44

Hence the TWRR, i, is given by

(1 + i)3 =x

2.2× 4.2x + 1.44

=21x

11(x + 1.44)The MWRR, i, is given by 4.2 = 2.2(1 + i)3 + 1.44(1 + i) or 55i3 + 165i2 + 201i− 14 = 0. Using the approximation(1 + i)3 ≈ 1 + 3i leads to i ≈ 0.07 and the true value will be less than this approximation.

Let f (i) = 55i3 + 165i2 + 201i− 14. Then f (0.06) = −1.33412 and f (0.07) = 0.897365. Using linear interpolationgives (0.07− i)/0.01 = 0.897365/(0.897365 + 1.33412) and hence i = 0.66.Using this value in the formula for the TWRR gives x = 11×1.44(1+i)3/

[21− 11(1 + i)3

]= 2.5000 or £2.5 million.

6. Use units of £1,000.Time, t 1 Jul 2003 1 Jul 2004 1 July 2005 1 Jul 2006Fund value at time t− 0 24 32 38Net cash flow at time t 5 8Fund value at time t + 0 21 29 40

(i) The MWRR, i, is given by 21(1 + i)3 + 5(1 + i)2 + 8(1 + i) = 38 leading to 21i3 + 68i2 + 81i − 4 = 0. Using theapproximation (1 + i)3 ≈ 1 + 3i and (1 + i)2 ≈ 1 + 2i leads to i = 0.049 and the true value will be less than this.Let f (i) = 21i3 + 68i2 + 81i− 4. Then f (0.049) = 0.1347386 and f (0.04) = −0.649856. Using linear interpolationgives: (0.049− i)/0.009 = 0.1347386/(0.1347386 + 0.649856) leading to i = 0.04745 or 4.75%.(ii) The TWRR, i, is given by

(1 + i)3 =2421× 32

29× 38

40=

1,2161,015

leading to i = 0.062078 or 6.21%.(iii) The return in the final year is poor (40 decreases to 38), whilst the returns in the first two years are good. TheMWRR is biased towards the return in the final year because the size of the fund is much larger in the final year.

7. Use units of £1,000,000.Time, t 1 Jan 2003 1 Jan 2004 1 July 2004 1 Jan 2005Fund value at time t− 0 450 500 800Net cash flow at time t 40 100Fund value at time t + 0 600 490 600

(i) (a) The TWRR, i, is given by

(1 + i)2 =450600× 500

490× 800

600=

5049

leading to i = 1.01015254 or 1.015%.

(i) (b) Effectively we have the following information:


Time, t 1 Jan 2003 1 Jan 2004 1 July 2004 1 Jan 2005Fund value at time t− 0 450 800Net cash flow at time t 40 100Fund value at time t + 0 600 490 600

For year 1 we have 1 + i1 = 450/600 = 3/4. For year 2, we have i2 with 490(1 + i2) + 100(1 + i2)1/2 = 800, or49(1 + i2) + 10(1 + i2)1/2− 80 = 0. This quadratic in (1 + i2)1/2 leads to (1 + i2)1/2 = (−10±

√100 + 320× 49)/98 =

(−5±√

3945/49) = 1.17978. Hence 1 + i2 = 1.39188.Then (1 + i)2 = (1 + i1)(1 + i2) = 0.75× 1.39188 and hence i = 0.0217197, or 2.17%.

(ii) The fund performs well in the period 1 July 2004 to 31 Dec 2004, but not so well in the period 1 Jan 2004 to1 July 2004. The TWRR for the year is based on the returns in the first 6 months and in the second 6 months, andignores the fact that there are differing amounts in the fund in these two periods. The LIRR for the year is based ona single return for the whole year—and there is more invested in the fund in the final 6 months.

8. Use units of £1,000,000.Time, t 1 Jan 2000 1 Jan 2001 1 July 2001 31 Dec 2001Fund value at time t− 0 43 49 53Net cash flow at time t 4 2Fund value at time t + 0 40 47 51

Between 1/1/2000 and 1/1/2001 the fund grows from 40 to 43; between 1/1/2001 and 1/7/2001 the fund growsfrom 47 to 49; and between 1/7/2001 and 1/1/2002 the fund grows from 51 to 53. Hence the TWRR is given by(1 + i)2 = (43/40)× (49/47)× (53/51). Hence i = 0.07921 and the TWRR is 7.92%.

For (b), this is equivalent to only having the following information.Time, t 1 Jan 2000 1 Jan 2001 1 July 2001 31 Dec 2001Fund value at time t− 0 43 53Net cash flow at time t 4 2Fund value at time t + 0 40 47

LIRR: first year. 40 grows to 43. Hence i1 = 3/40 = 0.075.LIRR: second year. 47 at time 0 and 2 at time 0.5 grow to 53 at time 1. Hence 47(1 + i2) + 2(1 + i2)0.5 = 53. Ifx = (1+ i2)0.5 then we get the quadratic 47x2 +2x−53 = 0. Hence x = (−1+2

√623)/47 and 1+ i2 = x2 = 1.083368.

Hence the LIRR is given by (1+i)2 = 1.075×1.083368 and so it is 7.92%. It is slightly less than the value of TWRR.(ii) The 2 values TWRR and LIRR will be equal if the same intervals are used. In this case, the 3 intervals 1/1/2000to 1/1/2001, 1/1/2001 to 1/7/2001 and 1/7/2001 to 1/1/2002 would have to be used for the calculation of the LIRR.Also, if the rate of growth in both subintervals is the same, then TWRR will be equal to LIRR. In this case, thegrowth in the first 6 months is 49/47 and is 53/51 in the second 6 months—these are very similar quantities and sothere is little difference between the values of TWRR and LIRR.

9. We have the following information.Time, t 1 Jan 2004 1 July 2004 1 Jan 2005 1 Jan 2006 31 Dec 2006Fund value at time t− 0 12.5× 1.05 = 13.125 20.9085 29.7225525 38.854229Net cash flow at time t 12.5 6.6 7.0 8.0Fund value at time t + 0 12.5 19.725 27.9085 37.7225525

(i) LIRR is given by (1 + i)3 = 1.05× 1.06× 1.065× 1.03. Hence i = 0.06879398 or 6.8794%.(ii) TWRR is given by

(1 + i)3 =13.125

12.520.908519.725

29.722552527.9085

38.85422937.7225525

= 1.05× 1.06× 1.065× 1.03

This gives i = 0.06879398 or 6.8794%.(iii) MWRR is solution of

12.5(1 + i)3 + 6.6(1 + i)2.5 + 7(1 + i)2 + 8(1 + i) = 38.854229If f (i) denotes the left hand side, then f (0.06) = 38.86789. Then f (0.055) = 38.454467. Interpolating: (i −0.055)/(38.854229 − 38.454467) = 0.005/(38.86789 − 38.454467) leading to i = 0.0598 or 5.98%. (In fact,5.98353%.)(iv) The MWRR is lower because the fund has less money at the beginning when rates are higher. The values ofTWRR and LIRR are equal because the value of the fund is known at the times of all cash flows.

10. We have the following information:Time, t 1 Jan 2012 30 September 2012 31 December 2012Fund value at time t− 0 1.9 0.8Net cash flow at time t −0.9Fund value at time t + 0 1.3 1.0


(i) TWRR is given by

1 + i =1.91.3× 0.8

1.0=

19× 8130

=7665

Hence TWRR = 0.16923 or 16.92%.(ii) The MWRR is the solution of 1.3(1+i)−0.9(1+i)1/4 = 0.8 or 13i+5−9(1+i)1/4. So let f (i) = 13i+5−9(1+i)1/4.Then f (0.3) = 8.9−9×1.31/4 = −0.71011; f (0.4) = 10.2−9×1.41/4 = 0.90030 and f (0.35) = 9.55−9×1.351/4 =−0.15121. By linear interpolation we have i− 0.35 = 0.15121× (0.05/1.05151) and so i = 0.35719 or 36% to thenearest 1%.(iii) The fund does well before the large withdrawal. It performs poorly after the large withdrawal when the size ofthe fund is small—this leads to a higher MWRR.

11. (i) (2.9/2.3) × (4.2/4.4) = 1.203557 or 20.36%. (ii) We want i with 2.3(1 + i) + 1.5(1 + i)8/12 = 4.2. So letf (i) = 23(1 + i) + 15(1 + i)2/3 − 42. Using the approximation (1 + i)2/3 ≈ 1 + 2i/3 gives i ≈ 4/33 = 0.12. Thenf (0.12) = −0.062803 and f (0.13) = 0.263347. Linear interpolation gives i = 0.12 + 0.01× 0.062803/(0.263347 +0.062803) = 0.121926 or 12.2%. (iii) The fund performs well from January to May when the amount in thefund is (relatively) small and performs badly from May to December—when the amount in the fund is large. Hencethe MWRR will be less than the TWRR, because the TWRR ignores the amount in the fund.

12. (a) Ignoring inflation we haveTime, t 1/4/2001 1/4/2002 1/4/2003 1/4/2006

Fund value at time t− 0 0 1000× 9080 1000

(1 + 90

80

) 9890 1000

[ 13066 + 473

36016098

]0 = 1125 = 2313.89 = 4114.82

Net cash flow at time t 1000 1000 0Fund value at time t + 0 1000 1000

(1 + 90

80

)1000

(1 + 90

80

) 9890

Hence the real TWRR is given by

(1 + i1)5 =9880×

13066 + 473

36016098(

1 + 9080

) 9890

× 11.22

=1048561

× 11.22

= 1.7856

This gives i1 = 0.12294104322 or 12.294% per annum.(b) The equation for the real MWRR is

1000 +1000ν1.02

= 1000[

13066

+473360

16098

]ν5

1.22= 4114.82

ν5

1.22or 19961ν5 − 5802.17647059ν − 5918.22 = 0. Substituting ν = 1/1.12294 gives LHS > 0. Hence MWRR >TWRR. (It is 12.50%.)

13. Use units of £1 million.Time, t 1 January 2010 1 January 2011 1 July 2012

Fund value at time t− 0 120 140 600Net cash flow at time t 0 200

Fund value at time t + 0 120 340

(i) Let i denote the annual time weighted rate of return. Then

(1 + i)5/2 =140120× 600

340=

3517

Hence i = 0.3348967 or 33.49%. (ii) The rate of growth in the first year is 16.67% but 46% in the last 18 months.Hence the rate of growth is higher when the fund contains more money; hence the money weighted rate of returnwill be higher than the time weighted rate of return. (It is 38.87%.)

14. (i) MWRR affected by amounts and timings of cash flows which are not under the control of the fund manager. Theequation for the MWRR may not have a unique or, indeed, any solution. But, TWRR requires knowledge of all thecash flows and the fund values at all the cash flow dates.(ii) Denote answer by i. Then (1 + i)2 = (45/41)× (72/19) leading to i = 0.17745183 or 17.75%.


1. (a) and (b) Use sn =∑n−1j=0 (1 + i)j , sn+1 =

∑nj=0(1 + i)j and sn =

∑nj=1(1 + i)j .

(c) If i = 0, an − an−1 = n− (n− 1) = 1. If i 6= 0, an − an−1 = (1 + i)(1− νn)/i− (1− νn−1)/i = 1.(d) If i = 0, ian + νn = 0 + 1 = 1. If i 6= 0, ian + νn = (1− νn) + νn = 1.(e) Use (d) and ian = dan . (f) If i = 0, LHS = RHS = 1. If i 6= 0, use equation 2.2a. (g) Use (f) and isn = dsnInterpretations: (b) (Value of n+ 1 payments at time n+ 1) = (value of first n payments at time n+ 1) + 1(for paymentat time n + 1). (c) (value of n payments at time 1) = (value of n − 1 payments at time 1)+1(payment at time 1).(d) Value at time 0 of loan of size 1, interest rate i with n annual interest payments. (e) As for (d), but interestpayments in advance. (f) and (g) correspond to (d) and (e) but valued at time n.


2. Now δ = 0.06; hence 1 + i = eδ and i = 0.0618. Either,

(Ia)10 =a10 − 10ν10

δ=ia10 ,i/δ − 10ν10

δ=ia10 ,i

δ2 − 10ν10

δ= 33.86

or, from first principles:

(Ia)10 =∫ 10

0tνt dt =

tνt

log ν

∣∣∣∣10

0−∫ 10

0

νt

log νdt =

10ν10

log ν− νt

(log ν)2

∣∣∣∣10

0= −10ν10

δ+

1− ν10

δ2 = 33.86

3. Let i denote the annual effective interest rate. Then 1 + i = 1.062. Let x = 1/(1 + i)1/4 = 1/√

1.06.

NPV = 40(x + x2 + · · · + x16) = 40x1− x16

1− x=

40√1.06− 1

(1− 1

1.068

)= 504.1267

Alternatively, use 2.3bwith i(4) = 4× (√

1.06− 1) = 0.118252. Hence NPV = 160a(4)4 ,0.1236 = 160×

(0.1236/i(4)

)×

a4 ,0.1236 = 167.236a4 ,0.1236 = 504.13

4. s(12)12 = (1 + i)n+1/ma(m)

n = 1.0712+1/12a(12)12 = 1.0712+1/12 ×

(i/i(12)

)× a12 = 18.5597

5. s(4)20 = (1 + i)20a(4)

20 = (1 + i)20.25 ×(i/i(4)

)× a20 = 1.07520.25 × 1.027701× 10.194491 = 45.316

6. The cash flow is as follows:Time 0 1 . . . 6 7Cash flow, c 200 190 . . . 140 130

210 210 . . . 210 210−10 −20 . . . −70 −80

Hence NPV(c) = 210a8 − 10(Ia)8 . (ii) Using an = (1 + i)an , (Ia)8 = (1 + i)(Ia)8 and (Ia)8 = (an −nνn+1)/(1−ν) gives NPV(c) = 210(1 + i)a8 −10(1 + i)(Ia)8 = 1.07

[210a8 − 10(a8 − 8ν9)/(1− ν)

]= 1076.82.

7. We want smallest integer n with 100 ≤ 20an,0.15. Using tables shows that n = 10 is the least such integer.8.

Time 0 1 2 3 4 5 6 7 8 9Cash flow, c 10 12 14 16 18 20 22 24 26 2810a10 10 10 10 10 10 10 10 10 10 102(Ia)9 0 2 4 6 8 10 12 14 16 188a10 8 8 8 8 8 8 8 8 8 82(Ia)10 2 4 6 8 10 12 14 16 18 20

NPV(c) =∑9t=0 8νt + 2

∑9t=0(t + 1)νt. Answer is (II) and (III) only.

9. Let x denote a revised payment. Then NPV = 600a(4)n,i = 12xa(12)

n,i. Hence

x = 50a(4)n,i

a(12)n,i

=50

(1 + i)1/12

a(4)n,i

a(12)n,i

=50

(1 + i)1/12

i(12)

i(4) =50eδ/12

12(eδ/12 − 1)4(eδ/4 − 1)

= 1501− e−δ/12

eδ/4 − 1= 49.17243

So the answer is £49.17.10. Using k2 = (k − 1)2 + 2k − 1 we can decompose the given cash flow c into c1 + c2 + c3 + c4 as follows:

Time 0 1 2 · · · k · · · n n + 1Cash flow, c 0 12 22 · · · k2 · · · n2 0Cash flow, c1 0 0 12 · · · (k − 1)2 · · · (n− 1)2 n2

Cash flow, c2 0 0 0 · · · 0 · · · 0 −n2

Cash flow, c3 0 2 4 · · · 2k · · · 2n 0Cash flow, c4 0 −1 −1 · · · −1 · · · −1 0

If x denotes the required answer, then x = νx− n2νn+1 + 2(Ia)n − an .11. The cash flow is as follows:

0 1 2 3 4 5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 120 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100

Let i1 = 0.04. ThenA(12) = 100

[(1 + i1)22 + (1 + i1)20 + (1 + i1)18 + (1 + i1)16 + (1 + i1)14 + (1 + i1)12]

+ 100[(1 + i1)11 + (1 + i1)10 + · · · + 1

]= 100(1 + i1)12 [(1 + i1)10 + (1 + i1)8 + · · · + (1 + i1)2 + 1

]+ 100s12 ,0.04

=100(1 + i1)12

(1 + i1) + 1s12 ,0.04 + 100s12 ,0.04 = 100

(1.0412

2.04+ 1)

s12 ,0.04 = 2681.835

OR, let x = 1/1.04. Then NPV = 100[x2 + x4 + · · · + x12 + x12a12 ] = 100[x2(1 + x2 + · · · + x10) + x12a12 ]. Hence

NPV = 100[x

1 + x(x + x2 + · · · + x12) + x12a12 ] = 100a12

x + x12 + x13

1 + x=

100a12

1.0412

1.0412 + 2.042.04


12. Let α = 1.0056 and β = 1.03. Thus α is the 6-month growth factor for t ∈ [0, 15] and β is the 6-month growth factorfor t ∈ (15, 25]. Clearly there are 50 payments in total.

The first 30 payments are made at times 0, 0.5, 1, . . . , 14.5. Their accumulated value at time 15 is c0 = 400(α30 +α29 + · · · + α); and their accumulated value at time 25 is β20c0.The last 20 payments are made at times 15, 15.5, 16, . . . , 24.5. The accumulated value of these 20 payments attime 25 is c1 = 400(β20 + β19 + · · · + β). Hence

A(25) = β20c0 + c1 = 400[β20α

α30 − 1α− 1

+ ββ20 − 1β − 1

]= 46,702.66

13. (i) Using first principles is probably the easiest method. We have the following cash flow:Time 0 1/12 2/12 3/12 · · · 64/12 65/12 66/12Cash flow 1/12 1/12 1/12 1/12 · · · 1/12 1/12 0

Let α = 1.131/12. Then

A(66/12) =1

12(α66 + α65 + · · · + α) =

α

12α66 − 1α− 1

=1.131/12

121.1311/2 − 11.131/12 − 1

= 7.882864

Or, i(12) = 12(1.131/12 − 1) and s(12)5.5 = 1.135.5a(12)

5.5 where a(12)5.5 = 1.131/12a(12)

5.5 = 1.131/12(1 − 1/1.135.5)/i(12).

Hence answer is 7.88286. Or, use (1 − d(12)/12)12 = (1/1.13) to get d(12) = 0.121597 and s(12)5.5 = is5.5/d

(12) =((1 + i)5.5 − 1)/d(12) = (1.135.5 − 1)/d(12) = 7.88286(ii) It is the accumulation after 51/2 years of an annuity with payments of size 1/12 each month in advance at aneffective interest rate of 13% per annum.

14. The first 2 results are in the notes. Then use the first result and s(m)n = (1 + i)na(m)

n and sn = (1 + i)nan to get thethird result. For the third result, use sn = (1 + i)sn and d(1 + i) = i.

15. (a) Just use an =∑nj=1 ν

j . (b) We want x with 10,000 = xa120 = x(a100 + ν100a20 ). Then use tables to getx = 143.47. Or work from first principles.

16.Time 0 1 2 3 4 5Cash flow 0 1 1 1 1 1

s5 = A(5) = 1 + exp(∫ 5

4 0.02t dt) + exp(∫ 5

3 0.02t dt) + exp(∫ 5

2 0.02t dt) + exp(∫ 5

1 0.02t dt).Now

∫ 5k

0.02t dt = 0.01(25− k2). Hence s5 = 1 + e0.09 + e0.16 + e0.21 + e0.24 = 5.773.Or, ν(t) = exp(−

∫ t0 δ(s) ds) = exp(−

∫ t0 0.02s ds) = exp(−0.01t2). Hence a5 = ν(1) + ν(2) + ν(3) + ν(4) + ν(5) =

e−0.01 + e−0.04 + e−0.09 + e−0.16 + e−0.25. Finally, s5 = a5/ν(5) = a5 e0.25 = 5.773

17. (a) NPV(c) =∑∞i=0(1 + g)k/(1 + i)k+1 = 1/(i− g) (b) 10,000/0.06 = 166,667

18. Just use limm→∞ i(m) = δ and limm→∞ d(m) = δ. See example 4.5c for this last limit.

19. If i = 0, then isn = 0 = dsn . If i 6= 0, then isn = (1 + i)n − 1 = dsn .If i = 0, then ian = 0 = dan . If i 6= 0, then ian = 1− νn = dan .The other results are in exercise 14.

20. LHS = a(m)n,i = (1 + i)1/ma(m)

n,i = (1 + i(m)/m)a(m)n,i = (1 + i(m)/m)(i/i(m))an,i = RHS

21. If i = 0 then LHS = n = RHS. If i 6= 0, then

RHS =1− νnm1

jm=

1− νn

i(m) = LHS where j =i(m)

mand ν1 =

11 + j

22. Let ν = 1/(1 + i) denote the discount factor. Using ν = 1− d = (1− dep)m where dep = d(m)/m gives

a(m)∞ =

1m

(1 + ν1/m + ν2/m + · · ·

)=

1m

11− ν1/m =

1mdep

=1d(m)

Using (1 + iep)(1− dep) = 1 gives iep = dep/(1− dep) and so i(m) = d(m)/(1− dep) = d(m)/v1/m. Hence

a(m)∞ =

1m

(ν1/m + ν2/m + · · ·) = ν1/m a(m)∞ =

ν1/m

d(m) =1i(m)

(b) follows immediately from the two tables of cash flows.

23. (a) Algebraically:

NPV(c) =k∑j=1

νjr = νr1− νkr

1− νr=

i

(1 + i)r − 1akr =

akrsr

Or: recall that sr denotes the accumulated value at time r of the sequence of r equal payments of 1 at times1, 2,. . . r. Hence the single payment of 1 at time r is equivalent to the r payments of 1/sr at times 1, 2, . . . ,r.Hence our annuity is equivalent to kr equal payments of 1/sr at times 1, 2,. . . , kr.(b) Set r = 1/m and k = nm (and hence kr = n) in previous result.


24. In the following x = ν1/m and y = ν1/q .

(I (q)a)(m)∞ =

1mq

[(1 + ν1/m + · · · + ν1/q−1/m) + 2(ν1/q + ν1/q+1/m + · · · + ν2/q−1/m) + · · ·

]=

1mq

[(1 + x + · · · + xm/q−1)(1 + 2y + 3y2 + · · ·)

]=

1mq

1− xm/q

1− x1

(1− y)2 =1mq

1− ν1/q

1− ν1/m

1(1− ν1/q)2 =

[1m

11− ν1/m

] [1q

11− ν1/q

]= a(m)∞ a(q)

∞ =1d(m)

1d(q)

25. Here we show the first equality. The second is done in exercise 35.

(Ia)n =∫ n

0ρ(t)ν(t) dt =

n∑j=1

∫ j

j−1jνt dt =

n∑j=1

j(νj − νj−1)log ν

=1 + ν + ν2 + · · · + νn−1 − nνn

δ

=an − nνn

δ

26. (a) Just use k|a(m)n = νka(m)

n and ν1/ma(m)n = a(m)

n .(b) If i = 0, then RHS = (n + k) − k = n = LHS. If i 6= 0, RHS = (1 − νn+k)/i(m) − (1 − νk)/i(m) =νk(1− νn)/i(m) = LHS.(c) Just use a(m)

n+k = (1 + i)1/ma(m)n+k and part (b).

27. Use the following decomposition:Time 0 1 2 · · · nCash flow, c 0 n n− 1 · · · 1(n + 1)an 0 n + 1 n + 1 · · · n + 1−(Ia)n 0 −1 −2 · · · −n

28. This decomposition gives (II):0 1 2 3 4 5 6 7 8 9 . . . 200 200 180 160 140 120 100 80 60 60 . . . 600 200 200 200 200 200 200 200 200 200 . . . 2000 0 -20 -40 -60 -80 -100 -120 -140 0 . . . 00 0 0 0 0 0 0 0 0 -140 . . . -140

This decomposition gives (III):0 1 2 3 4 5 6 7 8 9 . . . 200 200 180 160 140 120 100 80 60 60 . . . 600 60 60 60 60 60 60 60 60 60 . . . 600 140 120 100 80 60 40 20 0 0 . . . 0

The decomposition in (I) is incorrect:0 1 2 3 4 5 6 7 8 9 10 11 . . . 200 200 180 160 140 120 100 80 60 60 60 60 . . . 600 200 180 160 140 120 100 80 60 40 20 0 . . . 00 0 0 0 0 0 0 0 0 −60 −40 −20 . . . 0

60 60 60 60 60 60 60 60 60 60 60 60 . . . 60−60 −60 −60 −60 −60 −60 −60 −60 0 0 0 0 . . . 0

29. (i) We need i with

250 = 64a(4)5 ,i − 7a5 ,i =

64a(4)5

ν1/4 −7a5

ν= a5

[64(1 + i)1/4 i

i(4) − 7(1 + i)]

Using tables for a5 , (1 + i)1/4 and i/i(4) gives RHS = 253.87 for i = 0.05 and RHS = 248.371 for i = 0.06. Tryingi = 0.055 gives RHS = 251.096.Using interpolation between i = 0.055 and i = 0.06 gives (x−0.055)/(0.06−0.055) = (f (x)−f (0.055))/(f (0.06)−f (0.055)) and hence x = 0.055 + 0.005(250− 251.096)/(248.371− 251.096) = 0.057.Hence i = 0.057 or 5.7%.(ii) Using the general result that (1 + iR)(1 + e) = 1 + iM where iR is the real rate, e is the inflation rate and iM is themoney rate, we get iR = (i− e)/(1 + e) = (0.057− 0.02)/1.02 = 0.0363 of 3.63%.

30. First option: 456 = 240a(12)2 ,i . Using a(m)

n,i = id(m) an gives 240a(12)

2 ,0.05 = 240 × 1.026881 × 1.859410 = 458.254.Hence i > 0.05.Second option: 456 = 246a(12)

2 ,i . But 246a(12)2 ,0.05 = 246× 1.022715× 1.859410 = 467.805. Hence i > 0.05.

Both deals are unacceptable.


31. Let α = 1.03 and let ν = 1/1.04. First, note that∫ k

k−1νt dt =

νt

ln ν

∣∣∣∣kt=k−1

= νk−1 ν − 1ln ν

=iνk

δby using ν =

11 + i

and ln ν = −δ.

NPV of costs to companies plus NPV of costs to consumers is

60

[∫ 1

0νt dt +

∫ 2

1ανt dt + · · · +

∫ 20

19α19νt dt

]=

60iδ

[ν + αν2 + · · · + α19ν20]

=60iνδ

1− (αν)20

1− ανNPV of costs to financial advisers is

60∫ 1

0νt dt + 19

∫ 2

1νt dt + 18

∫ 3

2νt dt + · · · +

∫ 20

19νt dt =

i

δ

[60ν + 19ν2 + 18ν3 + · · · + ν20]

NPV of benefits to consumers is

30∫ 1

0νt dt + 33

∫ 2

1νt dt + · · · + 87

∫ 20

19νt dt =

i

δ

[30ν + 33ν2 + · · · + 87ν20]

NPV of benefits to companies is

12∫ 20

0νt dt = 12

νt

ln ν

∣∣∣∣20

t=0= 12

1− ν20

δ=

12iδ

a20

Hence

NPV =i

δ

[12a20 − 40ν + (10ν + 14ν2 + 18ν3 + · · · + 86ν20)− 60ν

1− (αν)20

1− αν

]Let S = 10ν + 14ν2 + 18ν3 + · · · + 86ν20. Then

(1− ν)S = 10ν − 86ν21 + 4ν2 1− ν19

1− νand so S =

10− 86ν20

i+ 4

1− ν19

i2=

86ia20 − 76 + 4a19

i

Hence

NPV =i

δ

[98a20 − 40ν +

4a19 − 76i

− 60ν1− (αν)20

1− αν

]= 1.019869×

[98× 13.590326− 40

1.04+

4× 13.133939− 760.04

− 601.04

1− (1.03/1.04)20

1− 1.03/1.04

]= −354.41

Net cost is £354.41 million.

32. Let ν = 1/1.04 and x = ν1/4. For the first alternative we have NPV = 300(1 + x + x2 + · · · + x39) = 300(1 −ν10)/(1−x) = 9975.208909 or £9,975.21. For the second alternative we have NPV = 2,520(ν2 +ν4 +ν6 +ν8 +ν10) =2,520ν2(1− ν10)/(1− ν2) = 10019.341845 or £10,019.34. Hence the first alternative is preferable for the borrower.

33. (i) Suppose α(t) =∫ t

0 δ(s) ds. For 0 ≤ t ≤ 4 we have α(t) = 0.08t. Note that α(4) = 0.32. For 4 < t ≤ 9 we haveα(t) = 0.32 + 0.12(t− 4)− 0.01t2/2 + 0.08 = −0.08 + 0.12t− 0.005t2. Note that α(9) = 0.595. For t > 9 we haveα(t) = 0.595 + 0.05(t− 9) = 0.145 + 0.05t. Hence


0δ(s) ds

)=

exp(−0.08t) for 0 ≤ t ≤ 4;exp(0.08− 0.12t + 0.005t2) for 4 < t ≤ 9;exp(−0.145− 0.05t) for t > 9.

We have ρ(t) = 100 exp(0.03t) for 10 < t < 12 and NPV(ρ) =∫ρ(u)ν(u) du =

∫ 1210 100 exp(0.03u) exp(−0.145 −

0.05u) du = 100 exp(−0.145)∫ 12

10 exp(−0.02u) du = 5,000(exp(−0.345)−exp(−0.385)) = 138.8485863 or 138.849.(iii) We want 1,000(e−0.08 + e−0.16 + e−0.24) = 2561.88799642 or £2,561.89.

34. For bond redeemed after 10 years:Time 0 1/2 2/2 · · · 19/2 20/2Cash flow after tax (in e) −100 3/2 3/2 · · · 3/2 3/2 + 100

The return is 1.5% after 6 months. Hence the effective rate of return is 1.0152 − 1 or 3.0225% p.a.For bond redeemed after 20 years:

Time 0 1/2 2/2 · · · 19/2 20/2 21/2 · · · 39/2 40/2Cash flow after tax (in e), c0 −100 3/2 3/2 · · · 3/2 3/2 21/8 · · · 21/8 21/8 + 130c1 −100 21/8 21/8 · · · 21/8 21/8 21/8 · · · 21/8 21/8 + 130c2 9/8 9/8 · · · 9/8 9/8 0 · · · 0 0

Note that c0 = c1 − c2. If i denotes the rate of return, then we need i with

100 =214

a(2)20 ,i + 130ν20 − 9

4a(2)

10 ,i


So let

f (i) = 100− 218x

1− x40

1− x− 130x40 +

98x

1− x20

1− x= 100− 130x40 − x

8(1− x)(12− 21x40 + 9x20)

where x = 1/(1 + i)1/2. Spreading the capital gain of 30 over 20 years gives 1.5% p.a. So the return is greater than3.0225 + 1.5. So try i = 0.045; this gives f (i) = −4.953363. The value i = 0.05 gives f (i) = 2.358334. Linearinterpolation gives i ≈ 0.045 + 0.005 × 4.953363/(2.358334 + 4.953363) = 0.048387 or 4.83% p.a. (Actually4.83377%.)(i) 4.83% p.a. (ii) 3.0225% p.a. (iii) Now 0.5(4.83377 + 3.0225) = 3.928 gives 3.928% p.a.(iv) The amount returned per e100 is

12

(3a(2)

10 ,i + 100ν10)

+12

(3a(2)

10 ,i +214ν10a(2)

10 ,i + 130ν20)

=(

3 +218ν10)

a(2)10 ,i + 50ν10 + 65ν20

=(

3 +218x20)x(1− x20)2(1− x)

+ 50x20 + 65x40

If i = 0.03928, the right hand side is 103.426 > 100; hence the answer is greater than 3.928% p.a. (It is actually4.21189% p.a.)

35. (i) Using ln ν = −δ and an =∫ n

0 νt dt = (1− νn)/δ gives

(Ia)n =∫ n

0ρ(t)ν(t) dt =

∫ n

0tνt dt =

nνn

ln ν−∫ n

0

νt

ln νdt =

nνn

ln ν+

1− νn

(ln ν)2

= −nνn

δ+

1− νn

δ2 =an − nνn

δ(ii) Amounts are in thousands. If ρ denotes revenue stream, then ρ(t) = 0 for t ∈ (0, 1); ρ(t) = 250(t − 1) fort ∈ (1, 11); ρ(t) = 2,500− 125(t− 11) for t ∈ (11, 29).Hence present value of revenue stream is

NPV(ρ) =∫ 29

1ρ(u)νu du =

∫ 11

1250(u− 1)νu du +

∫ 29

11ρ(u)νu du

= 250ν∫ 10

0uνu du + ν11

∫ 18

0(2500− 125u)νu du = 250ν(Ia)10 + ν11(2500a18 − 125(Ia)18 )

Using a18 = (1− ν18)/δ = ia18 ,i/δ = 0.2a18 ,0.2/ ln(1.02) = 5.2788 gives NPV(ρ) = 4761.094.Present value of opening cost and additional costs is 500a1 + 200a29 = 1548.470Hence answer is 4761.094− 1548.470 = 3212.624 or £3,212,624.

36. The cash flow is as follows:Time 0 1/12 2/12 3/12 4/12 5/12 6/12 7/12 8/12 9/12 · · · 24 11

12 25Costs −40 −3 −3 −3 −3 −3 −3Rental Income 1 1 1 1 · · · 1 0Maintenance − 2

12 − 212 − 2

12 · · · − 212 0

Sale 60After nmonths, where n > 8, we have NPV = −

[40 + 3(x + x2 + · · · + x6)

]+(x6+x7+· · ·+xn)− 1

6 (x7+x8+· · ·+xn).Hence

NPV = −40− 3x1− x6

1− x+ x6 1− xn−5

1− x− x7

61− xn−6

1− x=−40 + 37x + 17

6 x7 + x6 − 5

6xn+1

1− xHence NPV ≥ 0 when xn+1 ≤ 6

5

[37x + 17

6 x7 + x6 − 40

], or when n + 1 ≥ ln

[ 65

(37x + 17

6 x7 + x6 − 40

)]/ lnx =

112.55 where the inequality has been reversed because lnx < 0. Hence the DPP is 112 months, or 9 years and4 months.(b) The costs occur mainly in the first 6 months and the income is received later. If the interest rate decreases, thenthe NPV of the costs increases and the NPV of the income also increases. As the income is received later in timethan when the costs are incurred, the decrease in the interest rate will have a larger effect on the income. Hence theDPP will increase.If the interest rate was 0, then the NPV of the costs is -58 and the DPP would be month 76.

37. (i) We need k with 14a(2)k≥ 120; that means we need k with ak ≥ 60i(2)/7i = 8.42646. Tables shows that k = 14

is sufficient. Checking k = 13.5: a13.5 = (1− ν13.5)/i = 8.554839. Hence the answer is k = 13.5 years.(ii) The NPV of the first 13.5 years of payments is 14a(2)

13.5 = 14 ii(2) a13.5 . So profit is 14 i

i(2) a13.5 ,0.07 − 120 at 7%,which becomes 1.0713.51.0511.5

[14 i

i(2) a13.5 − 120]

at time t = 25.The value of the payments in the final 11.5 years at time t = 13.5 is 14a(2)

11.5 ,0.05 = 7.987092. The accumulated value

of these payments at time t = 25 is 1.0511.5 × 14a(2)11.5 ,0.05 = 213.3234.

So the total profit at time t = 25 is 221.3105.


38. We use units of £10,000. Let ν = 1/1.15 and x = ν1/4. NPV of costs:

500 + 350∫ 2

0νt dt = 500 + 350

νt

ln ν

∣∣∣∣2t=0

= 500 + 3501− ν2

ln 1.15= 1110.679238

NPV of benefits if sold after 5 years:

NPV =504

(x8 + x9 + x10 + x11) +554

(x12 + x13 + x14 + x15) +604

(x16 + x17 + x18 + x19) + 2050x20

=ν2

4(50 + 55ν + 60ν2)(1 + x + x2 + x3) + 2050v5 =

ν2

4(50 + 55ν + 60ν2)

1− ν1− x

+ 2050v5

=1

4× 1.155 (50× 1.152 + 55× 1.15 + 60)0.15

1− x+

20501.155 = 1122.03783


NPV =ν2

4(50)

1− ν1− x

+ 1650v3 =1

4× 1.153 × 50× 0.151− x

+16501.153 = 1120.80588


NPV =ν2

4(50 + 55ν)

1− ν1− x

+ 1800v4 =1

4× 1.154 (50× 1.15 + 55)× 0.151− x

+18001.154 = 1099.40298

(i) NPV criterion suggests selling after 5 years.(ii) DPP criterion suggests selling after 3 years.(iii) NPV of benefits if sold after 5 years (where x = 1/1.1751/4 and ν = 1/1.175):

NPV =1

4× 1.1756 (50× 1.1753 + 55× 1.1752 + 60× 1.175 + 65)0.1751− x

+X

1.1756

and this must equal

NPV(costs) = 500 + 3501− ν2

ln 1.175and hence

X = 1.1756(

500 + 3501− ν2

ln 1.175

)− (50× 1.1753 + 55× 1.1752 + 60× 1.175 + 65)

40.1751− x

= 2566.51624(iv) May be periods when property is unoccupied and rental income is reduced. Rental income may not increase atprojected rate. Tax and expenses (on rent and proceeds of sale). Maintenance and repair costs will increase as sale isdelayed. Forecast sale price is very high (£25,665,160)—this may not be achieved.

39. (i) (a) The equation NPV = 0 when considered as an equation for i. The NPV is the net present value of a sequenceof cash flows. (b) The DPP is the first time that the accumulated value of a project becomes positive.(ii) Inflows:

NPV(in) =∫ 6

3νt dt + 1.9

∫ 9

6νt dt + 2.5

∫ 15

9νt dt + 8ν15 =

ν6 − ν3

ln ν+ 1.9

ν9 − ν6

ln ν+ 2.5

ν15 − ν9

ln ν+ 8ν15

=ν3 − 1

ln ν[ν3 + 1.9ν6 + 2.5ν9(ν3 + 1)

]+ 8ν15

Outflows (where x = ν1/4 and α = 1/05):

NPV(out) = 1.5∫ 3

0νt dt +

0.34

[x12 + x13 + · · · + x59] + ν3[1 + αν + · · · + α11ν11]

= 1.5ν3 − 1

ln ν+ 0.075ν3 1− ν12

1− x+ ν3 1− (αν)12

1− ανUsing 9% gives NPV(in) = 12.62517 and NPV(out) = 13.32338. Hence the IRR is less than an effective 9% perannum.Now use 7% per annum on the first 12 years.

NPV(in) =∫ 6

3νt dt + 1.9

∫ 9

6νt dt + 2.5

∫ 12

9νt dt =

ν6 − ν3

ln ν+ 1.9

ν9 − ν6

ln ν+ 2.5

ν12 − ν9

ln ν

=ν3 − 1

ln νν3 [1 + 1.9ν3 + 2.5ν6] = 9.34598

NPV(out) = 1.5∫ 3

0νt dt +

0.34

[x12 + x13 + · · · + x47] + ν3[1 + αν + · · · + α8ν8]

= 1.5ν3 − 1

ln ν+ 0.075ν3 1− ν9

1− x+ ν3 1− (αν)9

1− αν= 12.55811

Hence the DPP is more than 12 years.


40. (i) Let α = 1.04 and ν = 1/1.11. Use units of £1,000. The net present value of the continuous income is24∑j=0

∫ j+1

j

80αjνs ds =24∑j=0

80αjνs

ln ν

∣∣∣∣s=j+1

s=j=

80ln ν

24∑j=0

αj(νj+1 − νj)

=80

ln ν(ν − 1)(1 + αν + α2ν2 + · · · + α24ν24) =

80ln ν

(ν − 1)1− α25ν25

1− αν= 968.241851

The net present value of the discrete payments is 500 + (100ν + 110ν2 + 120ν3 + 130ν4 + 140ν5) + 300ν15− 700ν25 =947.005379. Or, 500 + 90a5 ,0.11 + 10(Ia)5 ,0.11 + 300ν15 − 700ν25 where (Ia)5 ,0.11 = (a5 ,0.11 − 5ν6)/(1− ν).Hence the overall NPV of the project is 21.236472 or £21,236.47.(ii) The net present value of the outgoings is £998,531.04; the net present value of the income is £1019,767.51. Ifthe interest rate increases, then ν decreases and so the net present value of the outgoings and the income will bothdecrease. However, over half of the outgoings is the payment of £500,000 at time t = 0 which is unaffected by achange in ν. Hence the income will decrease by more than the outgoings and so the net present value of the projectwill decrease. For example, if ν = 1/1.12 then the NPV of the project is -£45,257.82.

41. Let ν = 1/1.08 and α = 1.012. In the first year, the total income is £(40,000× 365); hence the pension fund receives£(400 × 365) over the year. This amount is received continuously over the 365 days of the year. Measuring time inyears gives the net present value of the income from year 1 to be∫ 1

0ρ(t)ν(t) dt =

∫ 1

0(400× 365)νt dt = 400× 365

ν − 1ln ν

In the second year we have∫ 2

1ρ(t)ν(t) dt =

∫ 2

1(500× 1.1× 365)νt dt = 550× 365ν

ν − 1ln ν

Proceeding in this way gives the present value of the income is

365ν − 1ln ν

[400 + 550ν + 550αν2 + 550α2ν3 + · · · + 550α18ν19] = 365

ν − 1ln ν

[400 + 550ν

1− α19ν19

1− αν

]which is 2275323.80446 or £2,275,323.80. Hence the NPV is £275,323.80.

42. (i) Let i denote the internal rate of return and let ν = 1/(1 + i). Then NPV of outgoings is 1,309,500 + 12,000a(4)25 ,i.

The NPV of the incomings is

100,000a(4)5 ,i

[1 + 1.055ν5 + 1.0510ν10 + 1.0515ν15 + 1.0520ν20] = 100,000a(4)

5 ,i

1− 1.0525ν25

1− 1.055ν5

Hence i satisfies

13,095 + 120a(4)25 ,i = 1,000a(4)

5 ,i

1− 1.0525ν25

1− 1.055ν5

At 9%, the left hand side is 13095 + 120 × 1.033144 × 9.822580 = 14313. At 9%, the right hand side is 1000 ×1.033144× 3.889651× (1− 1.0525/1.0925)/(1− 1.055/1.095) = 14313.(ii) The NPV of the outgoings is 1,000,000. The NPV of the incomings is 85,000a(4)

20 ,i + 90,000a(4)5 ,i/(1 + i)20. Hence

the internal rate of return, i, satisfies f (i) = 1,000− 85a(4)20 ,i − 90a(4)

5 ,i/(1 + i)20 = 0. Now if there was no increase in

rent, i would satisfy 200/17 = a(4)25 ,i = (1 + i)1/4 i

i(4) a25 ,i which suggests i ≈ 0.08. Now f (0.08) = 43.217460361.Also f (0.07) = −39.0524598472. Using linear interpolation gives i = 0.07+0.01×39.0524598472/(43.217460361+39.0524598472) = 0.0747 and hence the answer is 7.5%.(iii) The answers to parts (i) and (ii) suggest project A is preferable, but:the outlay for project A is much larger than the outlay for project B—more money may have to be borrowed andinterest rates will change over the period of 25 years;the assumption of no increase in maintenance costs over 25 years for project A is questionable;the assumption of the rise in rents made for both projects A and B is questionable—-an attempt should be made toassess how likely each assumption will be met.

43. Let ν = 1/1.12, γ = ν1/4 and suppose α denotes the required compounding factor. The cash flow is as follows (unitsof $10,000):

Time 0 1/2 4/4 5/4 6/4 7/4 8/4 · · · 23/4 24/4Cash flow −400 −90 9 9 9 9 9α · · · 9α4 680

Hence 400 + 90ν1/2 = (1 + γ + γ2 + γ3)(9ν + 9αν2 + 9α2ν3 + 9α3ν4 + 9α4ν5) + 680ν6 = (1 + γ + γ2 + γ3)9ν(1 +αν +α2ν2 + α3ν3 + α4ν4) + 680ν6. Hence

1− α5ν5

1− αν=

1− ν1/4

9ν(1− ν)(400 + 90ν1/2 − 680ν6) = 4.55965787

The left hand side is a5 ,β where 1/(1 + β) = αν. Using an = an−1 + 1 gives a4 ,β = 3.55965787. Using tablesgives 0.045 < β < 0.05. Linear interpolation gives β = 0.0483515334 and hence α = 1.068344 or 6.83%.


44. Use units of £1,000. (i)(a) For project B we haveTime 0 1 2 · · · 19 20Cash flow—out -1,000Cash flow—in 60 60 60 · · · 60 1,000

Now 162/3× 60 = 1,000. Payments are in advance—so the payback period is 16 years.(i)(b) Let x = 1.005. Project A costs:

Time 0 1 2 · · · 19 20Initial cost −1,000Further costs −10 −10x −10x2 · · · −10x19 0

Project A income:Time 0 12/12 13/12 · · · 59/12 60/12 · · · 71/12 72/12 · · · 239/12 240/12Cash flow 0 5 5 · · · 5 5x · · · 5x 5x2 · · · 5x15 2,000

The income each year for project A is less than the income in the same year from project B. Hence the paybackperiod for project A will be longer.(ii)(a) The time when the present value of the incomings is equal or greater than the present value of the outgoings.(b) Project B: we need n with 60(1 + ν + · · · + νn) ≥ 1,000 or an+1 ,0.01 ≥ 50/(3×1.01). Using tables gives n = 18.(c) Again, the NPV of the income from project A in any year is less than that from project B in the same year; hencethe DPP for project A will be longer.(iii) If i denotes the IRR, then i satisfies 1,000 = 60a20 ,i + 1,000ν20. This leads to 1,000 = 60(1 − ν20)/(1 − ν) +1,000ν20 and hence ν = 94/100 and hence i = 6/94 = 0.063829787 or 6.383%.(iv) We work out the NPV of project A using the interest rate of 6.383% or i = 6/94 and ν = 94/100. For the costs wehave NPV = 1,000 + 10(1− x20ν20)/(1− xν) = 1122.868517. For the income we have

NPV = 60a(12)4 ν11/12 + 60xa(12)

1 ν59/12 + 60x2a(12)1 ,0.05ν

71/12 + · · · + 60x15a(12)1 ν227/12 + 2,000ν20

= 60a(12)4 ν11/12 + 60a(12)

1 xν59/12 [1 + xν + · · · + x14ν14] + 2,000ν20

= 60i

i(12)

[a4ν

11/12 + a1xν59/12 1− x15ν15

1− xν

]+ 2,000ν20

= 60i

12[(1 + i)1/12 − 1

] [1− ν4

iν11/12 + xν71/12 1− x15ν15

1− xν

]+ 2,000ν20 = 1227.154537

Hence NPV of project A at 6.383% is 1227.154537−1122.868517 = 104.286. Hence the IRR of project A is greaterthan 6.383%.(v) The IRR of project A is greater but the DPP is longer. But the DPP ignores the large final payment for project A.So the final decision will depend on other factors such as (a) comparison of NPV using various interest rate scenarios;(b) whether the capital would be useful for another project; etc.


1. Total repayment is 3 × £2,000 = £6,000. Hence original loan is £6,000 − £750 = £5,250. Hence the flat rate is750/(3× 5250) = 0.0476 or 4.8%.

2. The cash flow is

Time 0 1 2Cash Flow −1000 703.84 703.84

This leads to 1000 = 703.84(ν + ν2). The usual formula for the solution of a quadratic leads to ν = 0.792586 andhence i = 0.261693 or 26.17%.

3. Let x denote a payment. Then 5,000 = xa12 ,0.08. Using the prospective method, the capital outstanding after the 4thpayment is y = xa8 ,0.08. Hence the interest element of the 5th payment is 0.08y = 0.08 × 5000a8 ,0.08/a12 ,0.08 =305.02.

4. The monthly repayment, x, is given by 4,000 = x × 12a(12)5 ,0.15. Using a(m)

n = ii(m) an , see equation 2.3b, and tables

gives x = 4,000/(12× 1.067016× 3.352155) = 93.193 or £93.93.The flat rate of interest is (60x− 4000)/20000 = 0.0796 or 7.96%.

5. Let i denotes the effective annual rate of interest and let j denote the effective rate of interest over 1 month. Letν = 1/(1 + i); hence ν1/12 = 1/(1 + j). Then

5000 = 130× 12a(12)4 = 130(ν1/12 + ν2/12 + · · · + ν48/12) = 130a48 ,j

We need to solve a48 ,j = 5000/130 = 38.4615 for j. Tables give a48 ,0.005 = 42.580318 and a48 ,0.01 = 37.973959.Interpolating gives (x − 0.005)/(0.01 − 0.005) = (f (x) − f (0.005))/(f (0.01) − f (0.005)). Hence x = 0.005 +0.005(38.4615− 42.580318)/(37.973959− 42.580318) = 0.00947.Hence ν1/12 ≈ 1/1.00947 and so i ≈ 0.1197. Hence APR is 11.9% (nearest 0.1% rounded down).


6. Suppose the monthly repayment is x. Then 15,000 = 12xa(12)2 ,0.1236 = x(1 + α + α2 + · · · + α23) = x(1− α24)/(1− α)

where α12 = 1/1.1236. Hence x = 697.2717 which would be rounded up to £697.28. Hence flat rate is (24x −15,000)/30,000 = 0.057824 or 5.782%.

7. We need x with 100,000 = 12xa(12)20 ,j where j is the effective rate of interest per annum corresponding to a nominal

6% per annum convertible monthly.Now 6% per annum convertible monthly is 0.5% effective per month; so let ν = 1/1.005. Hence 100,000 =∑240k=1 xν

k = xa240 ,0.005 = xν(1− ν240)/(1− ν). Hence x = 716.43 or £716.43.(ii) After 6 years, 168 repayments are left. Hence capital outstanding is y = x(ν +ν2 + · · ·+ν168) = xν(1−ν168)/(1−ν) = x(1 − ν168)/0.005 = 81298.319786 or £81,298.32. Let ν1 = 1/(1 + 0.055/12) and let z denote the newrepayment. Then y = z(ν1 + ν2

1 + · · · + ν1681 ) = zν1(1− ν168

1 )/(1− ν1). Hence z = y(1− ν1)/ν1(1− ν1681 ) = 0.694960

or £694.96.

8. The annuity pays £10,000 per annum. Also i = 0.1.The capital outstanding after 5 years of payments is (prospective loan calculation) x = 10,000a(12)

15 ,0.1 = 10,000 ×1.045045 × a15 ,0.1 = 79,486.9588 or £79,487. Hence the interest component of the first instalment of the 6th yearis x×

(1.11/12 − 1

)= 633.84.

(ii) NPV of annuity over the whole 20 years is 10,000a(12)20 ,0.1 = 10,000 × 1.045045a20 ,0.1 = 88,970.57 or £88,971.

Hence capital repayment over the first 5 years is 88,971 − 79,487.34. Total payment of the annuity over the first5 years is £50,000. Hence total interest payment over the first 5 years is 50,000− (88,971−79,487.34) = 40,516.34.

9. Let x denote the quarterly payment. Then 20,000 = 4xa(4)20 ,0.1 = 4x×1.036756×a20 ,0.1 = 4x×1.036756×8.513564.

Hence x = 566.48.(ii) Loan outstanding after 24th payment is (prospective loan calculation) 4xa(4)

14 ,0.1 = 4x ii(4) a14 ,0.1. Hence the

interest element is[1.11/4 − 1

]× 4x i

i(4) a14 ,0.1 = xia14 ,0.1 = 417.31. Hence the capital element is x − 417.31 =149.17.

10. (i) We need x with 100,000 = 12xa(12)25 ,0.05 = 12x i

i(12) a25 . Hence x = 100,000/(12 × 1.022715 × 14.093945) =

578.138 or £578.14. (ii) (a) Using the prospective method, it is 12xa(12)13 ,0.05 = 12×578.14×1.022715×9.393573 =

66649.9311 or £66,649.93. (b) Interest portion of 145th payment is 66,649.93× (1.051/12 − 1) = 271.54. Hencecapital repayment is 578.14− 271.54 = 306.60.

11. (a) The annual interest payment is £5,000 × 0.1 = £500. (b) The annual sinking fund repayment, x is given byxs10 ,0.08 = 5,000. Hence x = 345.15. (c) The total of (a) and (b) is 845.15. The amortisation repayment, y is givenby ya10 ,0.1 = 5,000. Hence y = 813.73

12. (i) 50,000 = xa5 ,0.1 = x × 3.790787 and hence x = 13189.8732 or £13,189.87. Total interest = 5x − 50000 =15949.35. (ii) Capital outstanding at the beginning of the third year = the value at that time of all outstandingrepayments = xa3 ,0.1 = 32,801.25. We now want 32,801.25 = 4ya(4)

5 ,0.12 = 4y ii(4) a5 ,0.12 = 4y × 1.043938 ×

3.604776. Hence y = 2179.1012 and so the quarterly payment is £2179.10.(b) After first repayment, capital outstanding is z = y(ν1/4 + ν2/4 + · · ·+ ν19/4) = yν1/4(1− ν19/4)/(1− ν1/4). Henceinterest paid is z

[(1 + i)1/4 − 1

]= z(1− ν1/4)/ν1/4 = y(1− ν19/4) = 907.09 using ν = 1/1.12.

13. (i) We want x with 800,000 = xa10 ,0.08 and hence x = 800,000/6.710081 = 119,223.598 or £119,223.60. The totalinterest paid is 10x− 800,000 = 392,236.(ii) Capital outstanding at start of year 8 is xa3 ,0.08 = 119,223.60 × 2.577097 = 307250.7819. (a) We need ywith 307250.78 = 4ya(4)

5 ,0.12 = 4y ii(4) a5 ,0.12 = 4y × 1.043938 × 3.604776. Hence y = 20,411.7393 or £20,411.74.

(b) Interest payment is 307,250.78× [1.121/4 − 1] = 8,829.57. Hence capital repayment is 20,411.74− 8,829.57 =11,582.17.

14. The cash flow is as follows:Time 0 1 2 3 . . . 19 20Cash Flow −c 100 110 120 . . . 280 290

0 90 90 90 . . . 90 900 10 20 30 . . . 190 200

Hence c = 90a20 ,0.08 + 10(Ia)20 ,0.08 = 90a20 ,0.08 + 10(a20 ,0.08 − 20ν20)/0.08 = 90a20 ,0.08 + 10(1.08a20 ,0.08 −20ν20)/0.08 = 225a20 ,0.08 − 2500ν20 = 1672.71(ii) Using the prospective method, the capital outstanding after the 5th payment is equal to the value of all futurerepayments: hence l5 = 140a15 ,0.08 + 10(Ia)15 ,0.08 = 140a15 ,0.08 + 10(a15 ,0.08 − 15ν15)/0.08 = 140a15 ,0.08 +10(1.08a15 ,0.08 − 15ν15)/0.08 = 275a15 ,0.08 − 1875ν15 = 1762.78Hence the 6th repayment of 150 consists of interest 1762.78× 0.08 = 141.02 and capital 8.98.Hence the 7th repayment of 160 consists of interest (1762.78− 8.98)× 0.08 = 140.30 and capital 19.70.(iii) The last payment is £290. Let l19 denote the capital outstanding after the 19th payment. Hence the 20th payment


must consist of an interest payment of il19 and a capital repayment of l19. Hence (1+i)l19 = 290. Hence l19 = 268.519and 20th payment consists of an interest payment of 21.48 and a capital payment of 268.52.

15. Cash flow:

Time 1/1/80 1/2/80 1/3/80 1/4/80 . . . 1/12/09 1/1/10Loan −100,000Interest payments 0 500 500 500 · · · 500 500Investment premiums 1060/12 1060/12 1060/12 1060/12 · · · 1060/12 0

(a) NPV of investment premiums at 1/1/80 is 1060a(12)30 ,0.07 = 1060 i

i(12) a30 ,0.07 = 1060 ii(12) 1.071/12a30 ,0.07.

Accumulated value at time 1/1/10 is 1.0730NPV = 1060 ii(12) 1.071/12s30 ,0.07 = 1060 × 1.031691 × 1.071/12 ×

94.460786 = 103885.6856 or £103,885.69.

(b) Using effective 4% per annum gives 1060 ii(12) 1.041/12s30 ,0.04 = 1060 × 1.018204 × 1.041/12 × 56.084938 =

60730.42958 or £60,730.43. This implies a shortfall of £39,269.57.

(c) Revised premiums leads to additional accumulated value of 3940 ii(12) 1.041/12s10 ,0.04 = 48322.86400. This leads

to a surplus of £48,322.86-£39,269.57=£9,053.29.Or, total accumulated value is

1060i

i(12) 1.041/12s20 ,0.04 × 1.0410 + 5000i

i(12) 1.041/12s10 ,0.04

=i

i(12) 1.041/12 [1060s20 ,0.04 × 1.0410 + 5000s10 ,0.04]

=1.018204× 1.041/12 [1060× 29.778079× 1.0410 + 5000× 12.006107]

= 109053.2949as before.

(d) The interest rate on the investment is less than the rate on the loan. Hence the investor should have paid off theloan rather than investing in the policy.

(e) We assume monthly payments paid in arrears. The effective interest rate is iwith 1+i = 1.00512 and ν = 1/(1+i).Let ν1 = ν1/12.We need 12x where 100,000 = xa(12)

30 ,i = x[ν1 + ν2

1 + · · · + ν3601

]= xν1(1 − ν360

1 )/(1 − ν1). Hence 12x = 12 ×100,000× i/(1− ν30) = 6000/(1− ν30) = 7194.61

If we assume annual payments in arrears, we need x with 100,000 = xa30 ,i = xν(1 − ν30)/(1 − ν) = x(1 −ν30)/(1.00512 − 1). Hence x = 100,000(1.00512 − 1)/(1− 1/1.005360) = 7395.79.

16. (i) See derivation of equation 2.5b. (ii) The cash flow is as follows:

Time 0 1 2 3 . . . 20Cash flow, c 0 10 12 14 . . . 48

(a) Then NPV(c) = 10ν + 12ν2 + 14ν3 + · · · + 48ν20 = 402.372. or use NPV(c) = 8(ν + ν2 + · · · + ν20) + 2(ν + 2ν2 +· · · + 20ν20) = 8a20 + 2(Ia)20(b) The remaining cash flows are

Time 15 16 17 18 19 20Cash flow, c 0 40 42 44 46 48

Thus we want 40ν + 42ν2 + 44ν3 + 46ν4 + 48ν5 = 200.966. It is also 38a5 + 2(Ia)5 .(c) Now 0.03× 200.966 = 6.029. Hence the capital repayment is 40− 6.029 = 33.97.

17.Time 0 1 2 3 4 · · · 14 15Cash flow -c 50 48 46 44 · · · 24 22

Let ν = 1/1.06. Hence c = 50ν + 48ν2 + · · · + 24ν14 + 22ν15 = 2v[25 + 24ν + · · · + 12ν13 + 11ν14

]= 2νs where

s = 25 + 24ν + · · · + 12ν13 + 11ν14. Then (1− ν)s = 25− 11ν15 −[ν + ν2 + · · · + ν14

]. Hence

c =2i

[25− 11ν15 − 1− ν14

i

]=

1003

[25− 11ν15 − 50

3(1− ν14)

]= 370.5033

Hence size of loan is 370.5033.(ii) First payment: interest component is 370.5033× 0.06 = 22.230; hence capital repayment is 50− 22.23 = 27.77.Remaining loan outstanding is 342.7333.Second payment: interest component is 342.7333×0.06 = 20.564; hence capital repayment is 48−20.564 = 27.436.(iii) Using the prospective method: loan outstanding is 24ν + 22ν2 = 118,600/(53× 53) = 42.2214. Hence interestpayment of 14th instalment is 42.2214× 0.06 = 2.5333 and capital repayment is 24− 42.2214× 0.06 = 21.468.


18. The amount, x, of each monthly payment is given by

80,000 = 12xa(12)25 ,0.08 = 12x

i

i(12) a25 ,0.08 Hence x = 602.732.

(a) Interest at the end of the first month is 80000(1.081/12− 1) = 514.722. Hence the capital repaid is x− 514.722 =88.01. (b) Using the prospective method, the loan outstanding after the 228th payment is 12xa(12)

6 ,0.08. Hence the total

interest paid during the last 6 years is 72x−12xa(12)6 ,0.08 = 12x(6−a(12)

6 ,0.08) = 12x(6−ia6 ,0.08/i12)) = 8751.45. (c) The

capital outstanding, c is given by c×1.081/12 = 602.732; hence c = 598.88 and the interest paid is 602.73−598.88 =3.85. (ii) The total annual payments increase and the total interest paid increases.

19. (i) Let x denote the monthly payment which is paid in advance. Then 100,000 = 12xa(12)25 ,0.06 = 12x×1.061/12 i

i(12) ×a25 ,0.06. Hence x = 100,000/(12× 1.061/12 × 1.027211× 12.783356) = 631.5466 or £631.55.

Time 0 1/12 2/12 · · · 23/12 24/12 · · · 299/12 300/12 = 25Loan 100,000

Repayments x x x · · · x x · · · x 0

(ii) Use the prospective method. The outstanding loan at time 23/12 after 24 payments is 12xa(12)23 ,0.06 = 12x i

i(12) ×a23 ,0.06 = 95779.6067 or £95,779.61. Hence, interest component of next payment (the 25th) is 95,779.61 ×[1.061/12 − 1

]= 466.212 and the capital repayment is 631.55− 466.21 = 165.34.

(iii) (a) After 10 years at time 120/12, the loan outstanding is 12xa(12)15 ,0.06. Let y denote the new monthly pay-

ment. Then 12ya(12)15 ,0.02 = 12xa(12)

15 ,0.06 or y = xa(12)15 ,0.06/a

(12)15 ,0.02 = x ×

[1.061/12/1.021/12

]× a(12)

15 ,0.06/a(12)15 ,0.02 =

487.474900 or £487.47.(b) The reduction in instalments is x−y = 631.55−487.47 = 144.08 paid at times t = 120/12, 121/12, . . . , 299/12.We need the accumulated value of this cash flow at time t = 300/12 = 25. The value at time t = 119/12 is12ya(12)

15 ,0.02. Hence value at time t = 120/12 is 1.021/12 × 12ya(12)15 ,0.02 and the value at time t = 300/12 = 25 is

1.02181/12 × 12ya(12)15 ,0.02. In the usual way, a(12)

15 ,0.02 = ii(12) a15 ,0.02 = 1.009134× 12.849264. hence the value at time

t = 120/12 is 22455.810 and the value at time t = 300/12 is 30222.563.

20. (i) Now NPV = 1000[a10 ,0.05 + a10 ,0.07/1.0510

]= 12033.6051 or £12,033.61.

(ii) First note that 439.52/8790.48 ≈ 0.05; hence x ≤ 10. Loan outstanding at time t = k where 1 ≤ k ≤ 10 is1000

[a10 ,0.07/1.0510−k + a10−k ,0.05

]. Setting this equal to 8790.48 gives 8.79048 = 7.023582α10−k + 1−α10−k

0.05 =7.023582α10−k + 20(1 − α10−k) where α = 1/1.05. Hence 11.20952 = 12.976418α10−k; hence 10 − k =log(11.20952/12.976418) log(1/1.05) = 3. Hence k = 7 = x− 1 and so x = 8.(iii) Interest rate now remains at 5%. Value of annuity over 20 years is 1000a20 ,0.05 = 12,462.21, over 19 years is1000a19 ,0.05 = 12,085.321 and over 18 years is 11,689.587. Hence we need 18 payments of £1,000 with a reducedfinal payment. Denote this payment by x. Then NPV of the payments is 11,869.587 + x/1.0519 and we want this toequal 12,033.61 which implies a final payment of £869.32.Originally, total interest paid was £20,000-£12,033.61. Under the new scheme, interest paid is £18,000+869.32 -£12,033.61. The reduction is £2,000-£869.32=£1,130.68.

21. The cash flow is

Time 0 1 2 · · · 19 20Cash Flow −c 6000 5800 · · · 2400 2200

So c = 6000ν + 5800ν2 + · · · + 2400ν19 + 2200ν20 = 200x where x = 30ν + 29ν2 + · · · + 12ν19 + 11ν20 and(1− ν)x = 30ν − ν2(1− ν19)/(1− ν)− 11ν21 and so x = 42415.88.(ii) Cash flow after payment 7 is:

Time 7 8 9 · · · 19 20Cash Flow 0 4600 4400 · · · 2400 2200

Hence l7 = 4600ν + 4400ν2 + · · ·+ 2400ν12 + 2200ν13 = 200νy where y = 23 + 22ν + · · ·+ 12ν11 + 11ν12 = 27225.13.year loan outstanding repayment interest due capital repaid loan outstanding

before repayment after repayment8 l7 = 27225.13 x8 = 4600 il7 = 2450.26 x8 − il7 = 2149.74 l8 = 25075.399 l8 = 25075.39 x9 = 4400 il8 = 2256.79 x9 − il8 = 2143.21 l9 = 22932.18

(iii)After the ninth payment, the capital outstanding is l9 = 22932.18 and ν = 1/1.07.

Time 9 10 11 · · · 19 20Cash Flow −l9 x x− 200 · · · x− 1800 x− 2000

22932.18 = x(ν +ν2 + · · ·+ν11)− (200ν2 +400ν3 + · · ·+2000ν11) and so xν(1−ν11)/(1−ν) = 22932.18+200ν2(1+2ν + · · · + 10ν9) leading to x = 3924.09.


22. Let α = 1.05, ν = 1/1.06, ν12 = ν1/12 and ν3 = ν1/3 = ν412. The cash flow is as follows:

Time 1/9/98 1/7/99 1/11/99 1/3/00 1/7/00 1/11/00 · · · 1/11/03 1/3/04Cash Flow −c 1000 1000α 1000α2 1000α3 1000α4 · · · 1000α13 1000α14

Hencec = 1000

[ν10

12 + αν1412 + α2ν18

12 + α3ν2212 + α4ν26

12 + · · · + α13ν6212 + α14ν66

12

]= 1000ν10

12

[1 + αν3 + α2ν2

3 + α3ν33 + α4ν4

3 + · · · + α13ν133 + α14ν14

3

]= 1000ν5/6 1− α15ν5

1− αν1/3

and this is 17691.768 or £17,692.(ii) Loan outstanding on 1/7/99 is 17,692 × 1.0610/12 = 18,572.277 or £18,572.28. Hence interest repaid in firstinstalment is 18,572.28− 17,692 = 880.28 and so the capital repaid is 1000− 880.28 = 119.72.(iii) The seventh repayment is 1000α6 on 1/7/01. Capital outstanding at 1/3/01 (just after the 6th payment is made)is

1000α6ν412

[1 + αν4

12 + · · · + α8ν3212

]= 1000α6ν1/3 1− α9ν3

1− αν1/3 = 13341.566

or £13,341.57. Hence interest in 7th payment is 13341.57 ×[1.061/3 − 1

]= 261.67 and the capital repaid is

1000α6 − 261.67 = 1,078.43.

23. Let i denote effective interest rate per annum. Then 1 + i = 1.024. Let ν = 1/(1 + i)1/12 = 1/1.021/3. Let α = ν60 =1/1.0220. Then 10,000 = x(ν+ν2 + · · ·+ν60)+(x+10)(ν61 + · · ·+ν120)+(x+20)(ν121 + · · ·+ν180) which in turn equals[(1− ν60)/(1− ν)

] [10ν61 + 20ν121 + x(ν + ν61 + ν121)

]=[(1− α)/(1− ν)

]ν[10α + 20α2 + x(1 + α + α2)

]. It

follows that x(1 + α + α2) = 10,000(1 − ν)/ [(1− α)ν] − 10α − 20α2 and so x = 87.8345. hence the initialpayment will be set at £87.84.(ii) Interest at 1 May 1994 is 10,000(1.021/3 − 1). Hence capital repayment is 87.84− 10,000(1.021/3 − 1) = 21.61.(iii) Loan outstanding=(value at time k of original loan)-(accumulated value at time k of repayments). First term is10,000 × 1.0232. Let β = 1.021/3. Hence second term is x(β95 + β94 + · · · + β36) + (x + 10)(β35 + · · · + β + 1) =x(β96 − 1)/(β − 1) + 10(β36 − 1)/(β − 1) Hence answer is £6,708.31.(iv) Let revised amount be y. Then 6,708.31 = y(ν12 + ν24 + ν36 + ν48) = yν12(1 + ν12 + ν24 + ν36) = yν12(1 −ν48)/(1− ν12). Hence y == 2,036.35.

24. Now (Ia)n = ν + 2ν2 + · · · + nνn = ν(1 + 2ν + · · · + (n − 1)νn−1) = ν1−ν

[1−νn

1−ν − nνn]

= 1i

[an − nνn

]where

an is the present value of the annuity-due, a.(ii) Let ν = 1/1.08. Measuring in hundreds, x = 30ν + 32ν2 + 34ν3 + · · · + 58ν15. Hence (1 − ν)x = 30ν + 2ν2 +· · · + 2ν15 − 58ν16 = 30ν − 58ν16 + 2ν2(1− ν14)/(1− ν). Hence x = 35,255.57. Or use x = 28a15 + 2(Ia)15 .(iii) Let y denote loan outstanding at start of year 9. Then y = 46ν + 48ν2 + 50ν3 + · · · + 58ν7. As above, we havey = 1

1−ν

[46ν − 58ν8 + 2ν2(1−ν6)

1−ν

]= 267.5415.

Start of year 9: loan outstanding = £26,754.15. Repayment = £4,600. Interest = £26,754.15 × 0.08 = £2,140.33.Capital repayment is £2,459.67.Start of year 10: loan outstanding = £24,294.48. Repayment = £4,800. Interest = £24,294.48 × 0.08 = £1,943.56.Leaving loan outstanding = £21,438.04.(iv) Loan outstanding is now £21,438.04. Working in hundreds, we have:

Time using original origin 10 11 12 13 14 15Time using new origin 0 1 2 3 4 5Cash Flow −214.3804 x x + 2 x + 4 x + 6 x + 8

Hence

214.3804 = xν + (x + 2)ν2 + (x + 4)ν3 + (x + 6)ν4 + (x + 8)ν5 = xν1− ν5

1− ν+ 2ν2 [1 + 2ν + 3ν2 + 4ν3]

= x1− ν5

i+ 2ν2 [1 + 2ν + 3ν2 + 4ν3]

and hence

x =(214.3804− 2ν2 [1 + 2ν + 3ν2 + 4ν3])× i

1− ν5 = 47.12587

Hence the eleventh payment is £4,712.59.

25. First we find the monthly repayments.Let x denote the monthly repayment for loan A. Then (60x − 10000)/50000 = 0.10715. Hence x = 255.95833.Hence the repayment is £255.96.Let y denote the monthly repayment for loan B. Then

15,000 = 12y[a(12)

2 ,0.12 +1

1.122 a(12)3 ,0.10

]


Using the formula a(m)n = i

i(m) an and tables gives y = 324.4303 or £324.43. Hence answer to (i) is £600−£255.96−£324.43 = £19.61.(ii) First we find the effective interest rate per annum for loan A. We have 10,000 = 12 × 255.96 × a(12)

5 ,i . Hence

a(12)5 ,i = 3.255717 which leads to i = 0.2 or 20%. Hence capital outstanding under loan A is 12 × 255.96a(12)

3 ,0.2 =12× 255.96× 1.088651× 2.106481 = 7043.679 or £7,043.68.Capital outstanding under loan B is 12 × 324.43a(12)

3 ,0.10 = 12 × 324.43 × 1.045045 × 2.486852 = 10117.8255 or£10,117.83.So interest paid in 25th month is 7,043.68 × [1.21/12 − 1] + 10,117.83 × [1.101/12 − 1] = 188.5160 or £188.52.Hence, capital repayment is £255.96 + £324.43− £188.52 = £391.87.(iii) Total capital outstanding is £7,043.68 + £10,117.83 = £17,161.51. New monthly payment is £(255.96 +324.43)/2 = £290.20. Hence we need iwith 17161.51 = 12×290.20a(12)

10 ,i; hence we need iwith a(12)10 ,i = 4.9280697.

Using tables gives a(12)10 ,0.2 = 1.088651 × 4.192472 = 4.5641388 and a(12)

10 ,0.15 = 1.067016 × 5.018769 = 5.355107.Using linear interpolation gives i = 0.15 + 0.05 × (5.355107 − 4.9280697)/(5.355107 − 4.5641388) = 0.17699 oran effective rate of 17.7% per annum.Hence interest paid in 25th month is 17161.51 × [1.1771/12 − 1] = 234.6557 or £234.66. Capital repaid is290.20− 234.66 = 55.54 or £55.54.(iv) The new loan from Freeloans has a higher interest rate than loan B and a lower interest rate than loan A. So thestudent should investigate the possibility of using Freeloans to repay loan A only. Under the new arrangements, therepayments have been halved. Hence the student could afford to pay off the loan more quickly by maintaining thesame total repayments, if such a loan term could be negotiated with Freeloans.

26. Product A. We have 100,000 = 7,095.25a25 ,i and hence a25 ,i = 14.0939361. Hence i = 0.05 or 5% p.a.Product B.

27. (i)(a) The flat rate of interest isxR − l0nl0

=4800− 2000

2× 2000= 0.7

The flat rate of interest is 70%. (b) The flat rate of interest is not a good measure because it take no account of thefact that the loan outstanding decreases each month.(ii) The consumers’ association case.

Time 0 1/12 2/12 3/12 · · · 20/12 21/12 22/12 23/12Cash Flow −2,000 + 200 200 200 200 · · · 200 200 200 200

Hence

2000 = 2400a(12)2 ,i = 2400

i

d(12) a2 ,i or 5 = 6i

d(12) a2 ,i

Now if i = 2, then a2 ,i = (1− 1/9)/2 = 4/9 and (1− d(12)/12)12 = 1/3 and hence d(12) = 1.049823. Hence

6i

d(12) a2 ,i = 62

1.04982349

= 5.08 > 5

Hence the claim by the consumers’ association is correct.The banks’ case.

Time 0 1/12 · · · 11/12 12/12 · · · 17/12 18/12 · · · 23/12Receipts −2,000 + 200 200 · · · 200 120 · · · 120 60 · · · 60Costs −60 −60 · · · −60 −60 · · · −60 −60 · · · −60c1 60 60 · · · 60 60 · · · 60 60 · · · 60c2 60 60 · · · 60 60 · · · 60c3 80 80 · · · 80

Hence

2000 = c3 + c2 = 960a(12)1 ,i + 720a(12)

1.5 ,i =i

d(12)

[960a1 ,i + 720a1.5 ,i

]Now 1.01463× 1.025 = 1.40. At 4%,

i

d(12)

[960a1 ,i + 720a1.5 ,i

]= 1.021537

[960× 0.961538 + 720

1− ν1.5

0.04

]= 1993.51706 < 2000

Hence the banks’ case is also correct.

28. Work in units of £100. Let ν = 1/1.04.Time 0 1 2 3 4 · · · 11 12 13 · · · 19 20

−L 50 48 46 44 · · · 30 28 26 · · · 14 12c1 52 52 52 52 · · · 52 52 52 · · · 52 52c2 2 4 6 8 · · · 22 24 26 · · · 38 40

From first principles:L = 50ν + 48ν2 + · · · + 14ν19 + 12ν20


L(ν − 1) = 12ν21 + 2ν2 1− ν19

1− ν− 50ν

and hence

L =2ν(25− 6ν20)

1− ν− 2ν2(1− ν19)

(1− ν)2 = 456.386946

Alternatively

L = c1 − c2 = 52a20 − 2(Ia)20 = 52a20 − 2a20 − 20ν21

1− ν= 456.386946

Giving the answer £45,638.69.(ii) The 12th instalment is £2,800. Immediately after the 11th instalment has been paid at t = 11, the capital outstand-ing is the sum of all future repayments which is L1 = 28ν + 26ν2 + 24ν3 + · · · + 14ν8 + 12ν9. Hence

L1 = 30a9 − 2(Ia)9 = 30a9 − 2a9 − 9ν10

1− ν= 152.586727

Hence the capital outstanding at t = 11 is £15,258.67, and the interest component of the 12th instalment is 15,258.67×0.04 = 610.35. The capital repayment is therefore £2,800− £610.35 = £2,189.65.(iii) The loan outstanding just after the 12th instalment is paid is £15,258.67− £2,189.65 = £13,069.02. So we wantk and x where x < 28 and

130.6902 = 28ak + xνk+1

Hence k = 5 and xν6 = 130.6902− 124.651016 = 6.039184. Hence x = 7.64149437.We have established that the remaining term of the revised loan is 6 years with 5 repayments of £2,800 and one finalrepayment of £764.15.(iii) The total of all repayments is 50 + 48 + 46 + · · · + 28 + 5× 28 + 7.6415 = 615.6415. Hence the total interest paidis 615.6415− 456.386946 = 159.254554 or £15,925.46.

29. (i) Denote the annual payment by x. Then 500,000 = xa10 ,0.09 which gives x = 77910.041327 or £77,910.04. Thetotal amount of interest is 10× £77,910.04− £500,000 = £279,100.40.(ii)(a) Loan outstanding at t = 7 is value at t = 7 of the remaining 3 instalments. This is xa3 ,0.09. Let y denotethe amount of the new quarterly payment. Then 4ya4 ,0.12 = xa3 ,0.09 giving y = 77,910.04 × 2.531295/(4 ×1.043938 × 3.037349) or £15,549.16. (ii)(b) Let ν denote the quarterly discount factor. Hence ν = 1/1.121/4.Capital outstanding after first quarterly instalment is y(ν + · · · + ν15); capital outstanding after second quarterlyinstalment is y(ν + · · · + ν14). Hence capital repayment in second quarterly instalment is yν15 and so interest contentis y(1− ν15) = 15,549.16(1− 1/1.1215/4) = 5383.411819 or £5,383.41.

30. (i) Let ν = 1/1.04. The cash flow of the repayments is as follows:Time 0 1 2 . . . 14 15Cash Flow, c 0 400 380 . . . 140 120c1 0 420 420 . . . 420 420c2 0 20 40 . . . 280 300

So we want

c = c1 − c2 = 420a15 ,0.04 − 20(Ia)15 ,0.04 = 420a15 ,0.04 − 20a15 ,0.04 − 15ν16

1− ν= 3052.64507035

or £3,052.65.(ii) Interest component is £3,052.65× 0.04 = £122.11. So capital repayment is £400− £122.11 = £277.89.(iii) The beginning of the ninth year is just after the payment of £260 has been made. So the capital outstanding is

260a7 ,0.04 − 20(Ia)7 ,0.04 = 260a15 ,0.04 − 20a7 ,0.04 − 7ν8

1− ν= 1099.178046207

or £1,099.18. So new loan is £549.59. Let x denote the size of the new first repayment.Time 0 1 2 . . . 9 10Cash Flow, c −549.59 x x + 2 . . . x + 16 x + 18c1 0 x− 2 x− 2 . . . x− 2 x− 2c2 0 2 4 . . . 18 20

Hence 549.59 = (x− 2)a10 ,0.08 + 2(Ia)10 ,0.08 leading to x = 2 + (549.59− 2× 32.686907)/6.710081 = 74.162495or £74.16.

31. (i)Time 0 1 2 3 . . . 9 10Cash Flow, c 0 00 300 400 . . . 1,000 1,100c1 0 100 100 100 · · · 100 100c2 0 100 200 300 . . . 900 1,000


Now

c = c1 + c2 = 100[a10 ,0.06 + (Ia)10 ,0.06

]= 100

[7.360087 +

7.360087− 10ν11

1− ν

]= 4432.249451415

or £4,432.25. (ii) Immediately after the sixth repayment (at time t = 6 + 0), the loan outstanding is 800ν + 900ν2 +1000ν3 + 1100ν4 = 700a4 ,0.06 + 100(Ia)4 ,0.06 = 3266.64. Hence the interest component of the seventh repaymentis £3266.64× 0.06 = £196.00 and the capital component is £800− £196 = £604.(iii) The loan outstanding after the seventh repayment is £3266.64− £604 = £2662.64. Let x denote the new annualpayment. Then 2662.64 = xa8 ,0.08. Hence x = 463.3386576 or £463.34.

32. (i) After 1 month. amount owing is 300,000×1.0851/12. Hence monthly interest payment is 300,000× (1.0851/12−1) = 2,046.45 leading to a total interest payment over the first 15 years of 180× 2,046.45 = 368361 or £368,361.00.Similarly, for the remaining 10 years, the total interest paid is 40× 3,090.66 = 123,626.40. Total is £491,987.40.(ii) Let i denote the effective annual rate of interest. Then 1 + i = 1.0452. Let ν = 1/(1 + i) and x = ν1/12.Let z denote the monthly contribution to the savings account. Then the NPV of the 180 monthly contributions isz(1 + x + x2 + · · · + x179) = z(1 − x180)/(1 − x) = z(1 − ν15)/(1 − x). We want this to have value £150,000 after15 years. Hence 1.04530z(1 − ν15)/(1 − x) = 150,000 and z = 150,000(1 − x)/(1.04530 − 1) = 399.369219 or£399.37.(iii) Value of savings after 15 years is 399.37 × (1.115 − 1)/(1 − 1/1.11/12) = 160395.465880 or £160,395.46. Sooutstanding loan is £139,604.54. Let y denote the new monthly repayment. Then 139,604.54 = 12ya(12)

10 ,0.07. Hencey = 139604.54/(12× 1.031691× 7.023582) = 1605.498842 or £1,605.50.

33. (i) In the notes. (ii) Clearly 30,000 = X(Ia)60 ,0.0125. Easier from first principles: 30,000 = X(ν + 2ν2 + · · · +60ν60) = Xν

[(1− ν60)/(1− ν)2 − 60ν60/(1− ν)

]. Hence X = 26.6222 or £26.62. (iii) Let i denote the

effective rate of interest per annum and let ν = 1/(1 + j) = 1/(1 + i)1/12. Then 30,000 = 958.32ν36a60 ,j andhence ν36a60 ,j = 31.3047834. Using tables shows LHS = 31.4201977 when j = 0.01. So try j = 0.011; thenLHS=29.5097845. Linear interpolation gives j = 0.010060 and hence i = 0.0128 or 12.8%. (iv) The totalrepayments are 1,830X = 48,714.60 under the first schedule and 57,499.20 under the second schedule. So thesecond schedule has a lower interest rate but the total repayments are larger—the reason for this is that the paymentsdo not start for 3 years during which time the outstanding loan grows larger.

34. Let i denote the yield obtained by purchasing this loan for the price P . Then

P = gCa(m)n,i + Cανn = gC

1− νn

i(m) + Cανn =g

αi(m)

[C ′ −K

]+K where C ′ = Cα and K = C ′νn.

35. Use units of £1,000. Now K = value at time 0 of the instalments = 1.05× 20(ν4 + ν8 + ν12 + ν16 + ν20) = 21ν4(1−ν20)/(1− ν4) where ν = 1/1.08. Also C ′ = 100× 1.05 = 105. Finally, after allowing for tax, g = 0.1× 0.8 = 0.08and i(m) = 0.08(2) = 2(1.081/2 − 1). Hence P = 103.2856015262 or £103,285.60.


1. (a) £1,000,000(1 + 0.05× 90/365

)= 1,012,328.8;

(b) £1,012,328.8/(1 + 0.045× 67/365) = £1,004,035.2(c) Price on 4 May is £1,012,328.8/(1 + 0.04× 37/365). Hence we want[

price on 4 Mayprice on 4 April

− 1]

36530

=[

1 + 0.045× 67/3651 + 0.04× 37/365

− 1]

36530

= 0.050960034

or 5.1%.2. (a) A futures contract is a legally binding contract to buy or sell an agreed quantity of an asset at an agreed price at

an agreed time in the future. An option is a contract that gives the buyer the option of buying an agreed quantity ofan asset at an agreed price at an agreed time in the future. (b) Convertibles can be converted into ordinary sharesat a fixed price at a fixed date in the future. Hence they give the owner the option to purchase a specified quantity ofordinary shares at a specified price at a specified time in the future.

3. Suppose party A has a loan at a variable rate.Party A pays a series of fixed payments to party B for a fixed term.Party B, the other party to the interest rate swap, pays a series of variable payments—the interest payments on theloan.

4. Preference shares usually pay a fixed dividend whilst ordinary shares do not.Preference shares rank above ordinary shares for payments of dividends and if the company is wound up.Preference shares usually have a lower return to reflect the lower risk.Preference shares usually do not have voting rights; ordinary shares do.

5. Short term—less than one year. Sold at a discount.Used to fund short term spending of a government. In the UK, denominated in sterling, but also in euros.Sold at a discount and redeemed at par.Secure and marketable.Often used as a benchmark risk-free short term investment.


6. (i) Issued by the government. The term “gilt-edged market” refers to the market in government bonds. Coupon andredemption payments linked to an inflation index. (ii) Most bonds are linked to an inflation index but there is atime lag. When the bond is redeemed, the inflation is usually calculated on the basis of the inflation index at someearlier reference date. If inflation increases over the period of the bond, then the payments will not keep up withinflation.

7. (a) Market risk. Interest rates may move. If the current rate of 5.5% decreases, then the company receives less buthas to pay out the same each month. If the interest rate moves above 6%, the other party would have to pay out morethan it receives.

Credit risk: the other party defaults on the payments.

(b) At the current time, the company does not face a credit risk, because it is paying 6% but only receiving 5.5%.

8. Issued by companies. Entitle holders to share in profits (the dividend) after interest on loans, etc have been paid.They have lowest ranking. Higher expected returns than most other asset classes—but the risk of capital loss is alsohigher. Size of dividend is variable. Possible capital gain from increase in price of share. Marketability depends onsize of company. Voting rights in proportion to number held.

9. (a) Issued by large companies, banks and governments. Usually unsecured. Regular interest payments and redeemedat par. Issued in currency other than issuer’s home currency and outside the issuer’s home country. Yields depend onissuer, size of issue and time to redemption. Yields typically slightly lower than for conventional loan stocks fromthe same issuer. Usually a bearer security. Negotiable with an active secondary market.(b) Certificates issued by banks and building societies stating that money has been deposited. Usually last for 28 daysto 6 months. Interest paid on maturity. Security and marketability depends on issuing bank. Active secondary market.

10. (i) Purchase the security at time 0. This is a negative cash flow. Receive periodic coupon payments (usually every6 months or every year) and a capital repayment on maturity. The coupons (and possibly the capital payment) areindexed by a price index such as CPI or RPI. Often they are linked to the value of the index some time previously sothat there is no delay in making the payments.(ii) Purchase the share at time 0; this is a negative cash flow. Receive periodic dividends, usually every 6 months orevery year. Dividends are not fixed and depend on the success of the company. Dividends cease if company failsand then share may have no value. Receive value of share when it is sold—the value may be much higher than thepurchase price.

11. See notes on page 74 in section 4.4.

12. The running yield equals income divided by price. Some reasons:property is a less flexible investment;investing in property incurs high expenses (buying, selling and valuation expenses);difficult to value a property;commercial rents often fixed for 3 to 5 years but dividends can change every year.


1. The cash flow is as follows:Time 0 1/m 2/m · · · 1 (m + 1)/m · · · (nm− 1)/m n = nm/mCash flow, c −P fr/m fr/m · · · fr/m− t1fr fr/m · · · fr/m C + fr/m− t1fr

Hence i is a solution of the equation

P =fr

m

nm∑k=1

νk/m + νnC − t1frn∑k=1

νk

= fra(m)n,i + Cνn − t1fran

2. Define the function f by

f (n, i) = (1− t1)gCa(m)n,i + Cνn = C +

[(1− t1)g − i(m)]Ca(m)

n,i

Clearly f (n, i) as i. For bond A we have P = f (n1, i1) and for bond B we have P = f (n2, i2).(a) If P = C, then i(m)

1 = (1− t1)g = i(m)2 . Hence i1 = i2.

(b) If P < C then i(m)1 > (1 − t1)g. Also f (n, i1) as n . Hence f (n2, i1) < f (n1, i1) = P = f (n2, i2). But

f (n, i) as i. Hence i2 < i1.(c) If P > C then i(m)

1 < (1 − t1)g. Also f (n, i1) as n . Hence f (n2, i1) > f (n1, i1) = P = f (n2, i2). Usingf (n, i) as i shows that i2 > i1.


3. The cash flow is:

Time 0 t0 t0 + 1/m t0 + 2/m t0 + 3/m · · ·Cash flow −P (1− t1)fr/m (1− t1)fr/m (1− t1)fr/m (1− t1)fr/m · · ·

P =(1− t1)fr

m

(νt0 + νt0+1/m + νt0+2/m + · · ·

)=

(1− t1)frνt0

m

11− ν1/m = (1− t1)frνt0 a(m)

∞

Recall (1 − d(m)/m)m = 1 − d = ν and hence d(m) = m(1 − ν1/m). Hence an alternative expression is P =(1− t1)frνt0/d(m).

4. (a) Redemption is at the discretion of the bond issuer. A higher yield means that money is more expensive. This isgood news for the purchaser of the bond (low price P and high coupons) but bad news for a bond issuer. The currentyield on bonds is 10% p.a. and the coupons on the original bond are only 8% p.a. Hence, the bond issuer can makea profit by purchasing a new bond to pay the coupons on the existing bond rather than redeeming the existing bond.The bond issuer should retain the current bond—so the answer to part (a) is later.(b) Use equation 1.5c. We assume no tax. Hence P (n, i) = C + (g − i)Can,i = C + (0.08 − 0.06)Can,i =C[1 + 0.02an,i

]. Hence n implies i. Hence later redemption implies means a higher yield.

5. The cash flow is as follows (where t1 = 8/365):Time 0 t1 t1 + 1/2 t1 + 1 t1 + 3/2 t1 + 2 · · · t1 + 13/2 t1 + 7Cash flow, c −P 0 4 4 4 4 · · · 4 104

Because the bond is ex-dividend, the payment at time t1 is zero. The purchase price at time t1 should be 8a(2)7 +100ν7.

Using tables gives

P =1

1.068/365

[8a(2)

7 + 100ν7]

=8× 1.014782× 5.582381 + 66.5057

1.068/365 = 111.682

Alternatively, if x = 1/1.060.5, we have

P =1

1.068/365 4[x + x2 + · · · + x13 + x14 + 25x14] =

11.068/365 4x

[1− x14

1− x+ 25x13

]= 111.682

6. Now each coupon after tax is 350×0.75 = 262.50. If the bond is redeemed at the final possible date, the cash flow isTime 1/7/91 1/10/91 1/4/92 . . . 1/10/2009 1/4/2010Payment no. 0 1 2 . . . 37 38Cash flow −P 262.50 262.50 . . . 262.50 10,262.50

Now i(2) = 0.0591 and (1− t1)g = 0.75× 0.07 = 0.0525. As i(2) > (1− t1)g, assume bond will be redeemed at thelatest possible date (and then yield will be at least 6% whatever the redemption date). Let ν = 1/1.06. Hence

P = 262.5037∑k=0

ν0.25+0.5k + 10000ν18.75

= 262.50ν0.25 1− ν19

1− ν1/2 + 10000ν18.75 = 9385.46

(ii) Now i(2) = 2(1.051/2 − 1

)= 0.0494 and (1 − t1)g = 0.75 × 0.07 = 0.0525. As i(2) < (1 − t1)g, the second

purchaser should price the bond on the assumption it will be redeemed at the earliest possible date of 1/4/04 (andthen yield will be at least 5% whatever the redemption date). Hence the cash flow for the second purchaser is asfollows:

Time 1/4/99 1/10/99 1/4/00 . . . 1/10/2003 1/4/2004Payment no. 0 1 2 . . . 9 10Cash flow −P 262.50 262.50 . . . 262.50 10,262.50

Let ν = 1/1.05. Hence

P = 262.5010∑k=1

νk/2 + 10000ν5 = 262.50ν0.5 1− ν5

1− ν1/2 + 10000ν5 = 10136.30

7.Time (years) 0 1 2 . . . 19 20Cash flow −96 4 4 . . . 4 104

Hence 96 = 4a20 + 100ν20 and so 24 = a20 + 25ν20.Using method (d) in paragraph 4.3 gives i ≈

[4 + (100− 96)/20

]/96 = 0.044. If i = 0.04, RHS = 25.001; if i =

0.045,RHS = 23.374. Using interpolation gives (x−0.04)/(0.045−0.04) = (f (x)−f (0.04))/(f (0.045)−f (0.04))and hence x = 0.04 + 0.005(24− 25.001)/(23.374− 25.001) = 0.043. So answer is 4.3%.


8. Now i = 0.08 and so i(2) = 2[1.081/2−1] = 0.078461. Also (1−t1)g = 0.77×95/110 = 0.665. Hence i(2) > (1−t1)gand so CGT is payable. Hence P is given by

P = 0.77× 9.5a(2)20 ,0.08 + 110ν20 − 0.34(110− P )ν20

= 0.77× 9.5a(2)20 ,0.08 + (72.6 + 0.34P )ν20

and soP (1− 0.34ν20) = 0.77× 9.5a(2)

20 ,0.08 + 72.6ν20

which gives P = 95.79.

9. (i) Credit risk: most governments will not default. Low volatility risk. Income stream may be volatile relative toinflation. (ii) The cash flow is as follows:

Time (years) 0 1 2 3 4 5 6 7 8 9 10Cash flow, before tax −P 8 8 8 8 58 4 4 4 4 54

Let ν = 1/1.06. Then P = 8 × 0.7a5 + 50ν5 + 4 × 0.7a5ν5 + 50ν10 = (5.6 + 2.8ν5)a5 + 50ν5(1 + ν5) = 97.6855.

Or, P = 0.7× 4(a5 + a10 ) + 50ν5(1 + ν5) = 97.6855.

10. Now i(4) = 4[(1+ i)1/4−1] = 4[1.041/4−1] = 0.0394. Also (1−t1)g = 0.8×5/103 = 0.0388. Hence i(4) > (1−t1)gand so CGT is payable and the investor should assume latest possible redemption.Using units of £1,000 we have P = 5× 0.8a(4)

20 + 103ν20− 0.25(103−P )ν20 = 4a(4)20 + 77.25ν20 + 0.25Pν20. Hence

(1− 0.25ν20)P = 4a(4)20 + 77.25ν20 leading to P = 102.07201 or £102,072.01.

11. (i) Now t1 = 0.25, i(2) = 2(1.051/2 − 1) = 0.04939, and (1− t1)g = 0.75× 7/110 = 0.0477. Hence i(2) > (1− t1)g.Using the formula P (n, i) = C +

[(1− t1)g − i(m)

]Ca(m)

n,i shows there is a capital gain.(ii) Because i(2) > (1 − t1)g, it follows that the loan is least valuable to the investor if repayment is made by theborrower at the latest possible date.(iii) Assuming the loan is redeemed at the latest possible date, P = 0.75×3.5[x+x2+· · ·+x30]+[110−0.3(110−P )]x30

where x = 1/1.051/2. Hence P = 2.625x(1 − x30)/(1 − x) + [77 + 0.3P ]x30. Hence (1 − 0.3x30)P = 2.625x(1 −x30)/(1− x) + 77x30 giving P = 107.75380 or £107,753.80.Or, P = (5.25a(2)

15 + 77ν15)/(1− 0.3ν15) = 107.75383 or £107,753.83.

12. (i) The term “gross redemption yield” means before tax. Let ν = 1/1.05, then P = 4a(2)15 ,0.05 + 100ν15 = 4 ×

1.012348× 10.379658 + 100/1.0515 = 90.1330.(ii) (a) Now i(2) = 2[1.051/2 − 1] = 0.049390 and (1 − t1)g = 0.75 × 4/100 = 0.03. Hence i(2) > (1 − t1)gand CGT is payable. Now let ν = 1/1.06; then P = 4 × 0.75a(2)

7 ,0.06 + 100ν7 − (100 − P ) × 0.4ν7. Hence(1− 0.4ν7)P = 3a(2)

7 ,0.06 + 60ν7 leading to P = 77.5203.(b) The cash flow was as follows:

Time (years) 0 0.5 1 1.5 2 2.5 3 3.5 · · · 7.5 8Cash flow before tax −90.1330 2 2 2 2 2 2 2 · · · 2 2 + 77.5203Cash flow after tax −90.1330 1.5 1.5 1.5 1.5 1.5 1.5 1.5 · · · 1.5 1.5 + 77.5203

Let i denote the required effective rate of return per annum and let x = 1/(1+i)1/2. Then 90.133 = 3a8 ,i+77.5203ν8

and we need to solve for i.Using method (d) in paragraph 4.3 gives i ≈

[3 + (77.5203− 90.133)/8

]/90.133 = 0.016.

Define f (i) = 1.5x(1− x16)/(1− x) + 77.5203x16 where x = 1/(1 + i)1/2. Then f (0.015) = 91.3573 and f (0.02) =88.2486. Using linear interpolation gives (x − 0.015)/(90.133 − 91.3537) = (0.02 − 0.015)/(88.2486 − 91.3537)and hence x = 0.015 + 0.005 × 1.2207/3.1051 = 0.016966. Hence the return is 1.7% to the nearest 0.1%. (Moreprecisely, it is 1.694%.)

13. (i) Assuming £100 nominal is purchased, then price is

P1 = 0.75× 10a(2)10 ,0.08 +

1101.0810 = 102.26

Hence he paid £102.26 per £100 nominal.(ii) Now (1 − t1)g = 0.6 × 10/110 = 6/110 = 0.054 and i(2) = 2[(1 + i)1/2 − 1] = 2[1.061/2 − 1] = 0.059. Hencei(2) > (1− t1)g and so the investor is liable for CGT and the price is given by

P2 = 0.6× 10a(2)2 ,0.06 +

110− 0.4(110− P )1.062 = 6a(2)

2 ,0.06 +66 + 0.4P

1.062

and hence P2 = 108.544.(iii) We need to find i with P1 = 7.5a(2)

8 ,i + P2/(1 + i)8.Using method (d) in paragraph 4.3 gives i ≈

[7.5 + (P2 − P1)/8

]/P1 = 0.081.

Trying i = 0.08 gives RHS = 102.588. Trying i = 0.085 gives RHS = 99.689. Interpolating gives i ≈ 0.08 +0.005(f (i)− f (0.08))/(f (0.085)− f (0.08)) = 0.08 + 0.005× (102.26− 102.588)/(99.689− 102.588) = 0.0806 or8.1% approximately.


14. (i) The price is P = 5a20 ,0.06 + 100/1.0620 = 88.53.(ii) (a) Suppose the second investor pays P2. Now (1−t1)g = 0.7×8/100 = 0.056 and i = 0.065. Hence i > (1−t1)gand CGT is payable. Hence

P2 = 5× 0.7× a10 ,0.065 + [100− 0.3(100− P2)/1.06510

= 3.5a10 ,0.065 + [70 + 0.3P2]/1.06510

and so

P2 =3.5a10 ,0.065 + 70ν10

1− 0.3ν10 = 74.33

(b) We need i where 88.53 = 5a10 ,i + 74.33/(1 + i)10.Using method (d) in paragraph 4.3 gives i ≈

[5 + (74.33− 88.53)/10

]/88.53 = 0.0404. So try 4.5%: rhs =

87.4267. So must be lower: try 4%: rhs = 90.7692. Linear interpolation gives: (i − 0.045)/0.005 = (88.53 −87.4267)/(87.4267− 90.7692). Hence i = 0.04335 or 4.335%.

15. (i) Now t1 = 0.4 and g = 8/100 = 0.08. Also i(2) = 2(1.051/2 − 1) = 0.04939. Hence g(1 − t1) = 0.08 × 0.6 =0.048 < i(2). So assume redemption at the latest time of 2011. Also CGT is payable. Let ν = 1/1.05 and β = 0.6.

Time 1/1/01 1/7/01 1/1/02 . . . 1/7/2010 1/1/2011Cash flow −P 4β 4β . . . 4β 4β + 100− 0.3(100− P )

Hence, if x = ν1/2 = 1/1.051/2,

P = 0.6× 8a(2)10 ,0.05 +

70 + 0.3P1.0510

and so

P (1− 0.3x20) = 2.4x1− x20

1− x+ 70x20 and so P = 98.668

(ii) In this case, g = 0.08,t1 = 0 and i(2) = 2(1.071/2 − 1) = 0.06882. And so g(1 − t1) = 0.08 > i(2). So assumeredemption at the earliest time of 2006. Let ν = 1/1.07 and x = ν1/2 = 1/1.071/2.

Time 1/1/03 1/7/03 1/1/04 1/7/04 1/1/05 1/7/2005 1/1/2006Cash flow −P2 4 4 4 4 4 104

Hence P2 = 8a(2)3 ,0.07 + 100ν3 = 4x(1− x6)/(1− x) + 100x6 = 102.99.

(iii)Time 1/1/01 1/7/01 1/1/02 1/7/02 1/1/03Cash flow −P 2.4 2.4 2.4 2.4 + P2 − 0.3(P2 − P ) = 2.4 + 101.69

Hence 98.67 = 4.8a(2)2 ,i + 101.69ν2 = 2.4(x + x2 + x3 + x4) + 101.69x4 where x = 1/(1 + i)1/2. Using method (d) in

paragraph 4.3 gives i ≈[4.8 + (101.69− 98.67)/2

]/98.67 = 0.064.

Try i = 0.06. Then 98.67− 4.8a(2)2 ,i − 101.69ν2 = −0.764.

Try i = 0.065. Then 98.67− 4.8a(2)2 ,i − 101.69ν2 = 0.1353.

Linear interpolation gives (i− 0.06)/0.005 = (0 + 0.764)/(0.1353 + 0.764) and so i = 0.064248.Hence i(2) = 2× (1.0642481/2 − 1) = 0.0632.

16. We work in multiples of 100,000. The price per £100 nominal is

P1 = 8a(4)15 ,0.09 +

1101.0915 = 8

i

i(4) a15 ,0.09 +110

1.0915 = 96.821947

and hence £9,682,194.70.(ii) Now (1− t1)g = 0.8× 8/110 = 6.4/110 = 0.0581 and i(4) = 4[(1 + i)1/4 − 1] = 4[1.071/4 − 1] = 0.068. Hencei(4) > (1− t1)g and so CGT is paid. Hence the price is given by

P2 = 0.8× 8a(4)10 ,0.07 +

110− 0.2(110− P )1.0710 = 6.4

i

i(4) a10 ,0.07 +88 + 0.2P2

1.0710

which gives P2 = 101.131 and so answer is £101.131.(iii) Running yield is 100(1− t1)r/P2 = 0.8× 8/101.131 = 0.06328 or 6.382%.(iv) We need i with P1 = 8a(4)

5 ,i + P2/(1 + i)5. Using method (d) in paragraph 4.3 gives i ≈[8 + (P2 − P1)/5

]/P1 =

0.092. Trying i = 0.09 gives RHS = 97.877. Trying i = 0.095 gives RHS = 96.275. Interpolating givesi ≈ 0.09+0.005(f (i)−f (0.09))/(f (0.095)−f (0.09)) = 0.09+0.005×(96.822−97.877)/(96.275−97.877) = 0.093or 9.3% approximately.

17. (i) Now (1− t1)g = 0.75× 9/110 = 0.0613 and i(2) = 2[1.061/2 − 1] = 0.059. Hence i(2) < (1− t1)g and so there isno CGT. The price P1 is given by

P1 = 0.75× 9a(2)13 ,0.06 + 110ν13 = 112.21121

Hence answer is £112.21.(ii)(a) Now (1− t1)g = 0.9× 9/110 = 0.0736 and i(2) = 2[1.081/2− 1] = 0.0784. Hence i(2) > (1− t1)g and so there


is CGT for the second investor. The price P2 is given by

P2 = 0.9× 9a(2)11 ,0.08 +

110− 0.35(110− P2)1.0811 = 8.1

i

i(2) a11 ,0.08 +71.5 + 0.35P2

1.0811

which leads to P2 = 105.45468 or £105.45.(b) The cash flow for the first investor is as follows (note there is no capital gain):

Time 0 0.5 1 1.5 2Cash flow before tax −P1 4.5 4.5 4.5 4.5 + P2Cash flow after tax −P1 3.375 3.375 3.375 3.375 + P2

Then P1 = 6.75a(2)2 ,i + P2ν

2. Using method (d) in paragraph 4.3 gives i ≈[6.75 + (P2 − P1)/2

]/P1 = 0.030.

Let f (i) = P1 − 6.75a(2)2 ,i − P2ν

2. Then f (0.03) = −0.203 and f (0.04) = 1.854. Using linear interpolation gives(i− 0.03)/0.203 = 0.01/(1.854 + 0.203) and hence i = 0.031. So the answer is 3% to the nearest 1%.

18. Now i = 0.06 and hence i(4) = 4(1.061/4− 1) = 0.058695. Also (1− t1)g = 0.6× 100× 0.08/105 = 0.04571. Hencei(4) > (1 − t1)g and the purchaser should value on the assumption that redemption takes place at the latest possibletime. Also, capital gains will be paid. So we want P with

P = 4.8a(4)20 ,0.06 +

105− 0.3(105− P )1.0620 = 4.8

i

i(4) a20 ,0.06 +73.5 + 0.3P

1.0620

which gives P = 87.37 or £87,370.(a) Now (1− t1)g = 0.8× 8/105 = 0.060952 and i(4) = 0.058695. Hence i(4) < (1− t1)g and there is no CGT. Thevalue of the coupons and redemption payment at 12 months after the time of original issue is

x = 0.8× 2 + 0.8× 8a(4)14 ,0.06 + 105ν14 = 1.6 + 6.4

i

i(4) a14 ,0.06 + 105ν14 = 108.8517

where the term 0.8 × 2 is for the coupon at time 12 months. It follows that P = ν1/6x = 107.7997 and so the priceis £107,799.70.(b) Again, there is no CGT. The value of the coupons and redemption payment at 12 months after the time of originalissue is

x = 0.8× 2 + 0.8× 8a(4)19 ,0.06 + 105ν19 = 1.6 + 6.4

i

i(4) a19 ,0.06 +105

1.0619

and hence P = ν1/6x = 108.2467 and the price is £108,246.72.(ii) Purchaser should pay no more than 107,799.70. If purchaser pays more than this and bond is redeemed early,then yield will be less than 6%.

19. (i) Now t1 = 0.3 and g = 8/100 and hence (1 − t1)g = 0.056; also i(2) = 2(√

1.06 − 1) = 0.05913. Hencei(2) > (1− t1)g. Assume capital gain and bond redeemed at latest possible time.

Time 1/5/11 1/7/11 1/1/12 1/7/12 · · · 1/1/17 1/7/17 · · · 1/7/21 1/1/22Before tax −P 4 4 4 · · · 4 4 · · · 4 104After tax −P 2.8 2.8 2.8 · · · 2.8 2.8 · · · 2.8 2.8 +

75 + 0.25P

Let ν = 1/1.06. Then

P = 1.061/3(

5.6a(2)11 ,0.06 + (75 + 0.25P )ν11

)and hence

P =1.061/3 × 5.6a(2)

11 ,0.06 + 1.061/3 × 75ν11

1− 1.061/3 × 0.25ν11 = 99.3188735468

or £99.31887. (ii) Now t1 = 0, g = 8/100 and i(2) = 2(√

1.07 − 1) = 0.0688. Hence (1 − t1)g > i(2). Assumeno capital gain and bond is redeemed at earliest time of 1/1/2017.

Time 1/4/13 1/7/13 1/1/14 1/7/14 1/1/15 1/7/15 1/1/16 1/7/16 1/7/17−P1 4 4 4 4 4 4 4 104

Let ν1 = 1/1.07. Then

P1 = 1.071/4(

8a(2)4 ,0.07 + 100ν4

1

)= 105.624983612

or £105.6250. (iii) We haveTime 1/5/11 1/7/11 1/1/12 1/7/12 1/1/13 1/4/13Before Tax −99.31887 4 4 4 4 105.625After tax −99.31887 2.8 2.8 2.8 2.8 104.0485

Now 1.081/3 × (2.8/1.081/2 + 2.8/1.08 + 2.8/1.083/2 + 2.8/1.082) + 104.0485/1.0823/12 = 100.2255.Also 1.091/3 × (2.8/1.091/2 + 2.8/1.09 + 2.8/1.093/2 + 2.8/1.092) + 104.0485/1.0923/12 = 98.568.Hence the net yiell is between 8% and 9%.


20. Now (1 − t1)g = 0.75 × 7/108 = 0.04861 and i(m) = i(4) = 4(1.051/4 − 1) = 0.04909. Hence i(m) > (1 − t1)g andso there is a capital gain and assume late redemption. Let P denote the price per £100. Then P = 0.75× 7a(4)

20 ,0.05 +[108− 0.35(108− P )] /1.0520. Hence

P

[1− 0.35

1.0520

]= 5.25a(4)

20 ,0.05 +70.2

1.0520 leading to P = 107.24537545 and the answer £107,245.38.

21. (i) Let ν = 1/1.03158. Then P1 = 5a18 ,0.03158 + 100ν18 = 5ν(1− ν18)/(1− ν) + 100ν18 = 124.9988 or £125.00.(ii) Now let ν = 1/1.05. Then P2 = 5a13 ,0.05 + 100ν13 = 100.(iii) Let ν = 1/(1 + i) where i is the required gross annual rate of return. Then 125 = 5(ν + ν2 + ν3 + ν4 + ν5) + 100ν5.This implies ν = 1 and i = 0. Hence the rate of return is 0%.(iv) The value of P1 would be larger. Hence we would have ν > 1 in part (iii). This implies i < 0.

22. (i) Define i by 1 + i = 1.0152. Let i′ = 0.015.Time 0 1/2 2/2 3/2 · · · 19/2 20/2Cash flow −P 2 2 2 · · · 2 2

So P = 4a(2)10 ,i + 100/(1 + i)10 = 2a20 ,0.015 + 100/1.01520 = 108.584319822 or £108.58.

(ii) Assuming tax of 25%, the price at time 0 is P = 1.5a20 ,0.015 + 100/1.01520 and hence the price 91 days later isP × 1.015182/365 = 100.745153968 or £100.75.

23. (i) The return from bond A is i = 5/101. Let P denote the price of bond B then the expected return is[0.1× (−P ) + 0.2(100− P ) + 0.3(50− P ) + 0.4(106− P )] /P = [77.4− P ] /P

Hence we want [77.4− P ] /P = 5/101 which gives P = 77.4× 101/106 = 73.749057 or e73.749.(ii) Let i′ denote the gross redemption yield—the gross redemption yield always ignores the possibility of default.Hence 73.749(1 + i′) = 106 giving 1 + i′ = 1.437308 and hence 43.73%.(iii) Bond B has higher risk—hence an investor will require a higher return.


1. If the payments are linked exactly to the inflation index, then the real yield will be the same whatever happens toinflation. However, the delay of 8 months before allowance is made for inflation means that the investor will beworse off.

2. Using (1 + iM ) = (1 + q)(1 + iR) gives (1 + 0.01) = (1 − 0.02)(1 + iR). Hence 1 + iR = 1.01/0.98 and soiR = 3/98 = 0.3061 or 3.061%.

3. Let α = 1.03 and ν = 1/1.08. Then

P = 5∞∑k=1

αkνk/2 =5αν1/2

1− αν1/2 =5α

(1 + i)1/2 − α=

5.151.081/2 − 1.03

= 557.93 Hence £557.93.

4. Let α = 1.025 and ν = 1/1.07. Then

P = 2∞∑k=1

αkνk/2 =2αν1/2

1− αν1/2 =2α

(1 + i)1/2 − α=

2.051.071/2 − 1.025

= 217.90 Hence £217.90.

5. (i) Issued by a government; coupons paid by government—hence virtually no risk of default in most cases. Couponand redemption values linked to a specified inflation index, usually with a time lag. hence they provide protectionagainst inflation.(ii) Usually, payments linked to value of inflation index some months earlier. If inflation is higher over these months,then the payments will not keep up with inflation.

6. (a) If i denotes the money rate of return, then 10(1 + i) = 11.1 and hence i = 0.11 or 11%.(b) If q denotes the rate of inflation, then 112(1 + q) = 120 and hence q = 8/112 = 0.07143 or 7.143%.(c) Using (1 + 0.6i) = (1 + q)(1 + iR) gives 1 + iR = (1 + 0.6 × 0.11) × 112/120 and hence iR = −0.005067, or−0.5067%.

7. Let α = 1.04 and ν = 1/1.07. Then P = 12[ν2/12 + αν14/12 + α2ν26/12 + · · ·

]= 12ν1/6

[1 + αν + α2ν2 + · · ·

]=

12ν1/6/ [1− αν] = 428ν1/6 = 423.20

8. The cash flow is as follows:Time 0 1 2 3 4 5 . . .Cash flow 0 800 800(1.08) 800(1.08)(1.07) 800(1.08)(1.07)α 800(1.08)(1.07)α2 . . .

where α = 1.05. Let ν = 1/1.07. Hence required value is

NPV = 800ν + 800(1.08)ν2 +∞∑k=3

800(1.08)(1.07)αk−3νk = 800ν + 864ν2 +924.48ν3

1− αν= 41876.15

Giving an answer of £418.76.


9. Cash flow is as follows:Time 0 1 2 3Cash flow, unadjusted −20 2 2 22Inflation index, Q(t) 245.0 268.2 282.2 305.5Cash flow, adjusted −20 2×245.0

268.22×245.0282.2

22×245.0305.5

Hence

20 =2× 245.0ν

268.2+

2× 245.0ν2

282.2+

22× 245.0ν3

305.5=

490ν268.2

+490ν2

282.2+

5390ν3

305.5

Inflation index goes from 245.0 to 305.5 in 3 years. Setting (1 + j)3 = 305.5/245.0 gives j = 0.076 as the averagerate of inflation. General result (1 + iR)(1 + j) = 1 + iM where iR is the real rate, j is the inflation rate and iM is themoney rate, leads to iR = 0.022 as a first estimate of the real rate of return. (Or use approximation νk ≈ 1− ki.)Trying i = 0.022 gives ν = 1/1.022 and RHS = 19.978. Trying i = 0.021 gives ν = 1/1.022 and RHS = 20.032.Interpolating gives (x− 0.021)/(0.022− 0.021) = (f (x)− f (0.021)/(f (0.022)− f (0.021)) and hence x = 0.021 +0.001(20− 20.032)/(19.978− 20.032) = 0.0216 or 2.16%.

10. Effective real yield per annum,iR, is given by 1 + iR = 1.01252. Cash flow is:Time 0 10/12 22/12 34/12 · · ·Cash flow 0 5 5× 1.03 5× 1.032 · · ·Inflation Index 1 1.0210/12 1.0222/12 1.0234/12 · · ·

Hence, if β = 1 + iR = 1.01252, we have

x =∞∑n=0

5× 1.03n

1.02n+10/12(1 + iR)n+10/12 =5

(1.02β)10/12

∞∑n=0

1.03n

1.02nβn= 5

(1.02β)1/6

1.02β − 1.03

Hence x = 321.68 or £3.217.

11. Let νm = 1/(1 + iM ) where iM is the money rate of return. Hence

1 =∞∑k=1

d(1 + g)k−1νk =dν

1− (1 + g)ν=

d

iM − gHence iM = d + g. Using the general result that (1 + iR)(1 + e) = 1 + iM where iR is the real rate, e is the inflationrate and iM is the money rate, we get iR = (d + g − e)/(1 + e) as required.

12. Let α = 1.03 and β = 1.05. The cash flow is as follows:Time (in years) 0 0.75 1.75 2.75 3.75 . . .Cash flow, unadjusted −250 10 10β 10β2 10β3 . . .Inflation factor 1 α0.75 α1.75 α2.75 α3.75 . . .

Hence we require:

250 =∞∑k=0

10βkν0.75+k

α0.75+k =10ν0.75

α0.75

∞∑k=0

βkνk

αk=

10ν0.75

1.030.75

11− 1.05ν/1.03

and this leads to0 = 25× 1.05ν + 1.030.25ν0.75 − 25× 1.03 = 26.25ν + 1.030.25ν0.75 − 25.75

or, multiplying by 1 + i,

0 = 0.5 + 1.030.25(1 + i)0.25 − 25.75iWe need an initial approximation.Either, approximate (1 + i)0.25 by 1 + 0.25i. This leads to i ≈

(0.5 + 1.030.25

)/(25.75− 0.25× 1.030.25

)= 0.059.

In fact, (1 + i)0.25 = 1 + 0.25i + α where α > 0; hence i is greater than the approximation 0.059.Or, ignoring inflation and assuming the next dividend occurs in 12 months’ time gives 250 = 10ν + 10βν2 + · · · =10ν/(1 − βν) = 10/(1 + iM − β) = 10/(iM − 0.05). Hence iM ≈ 0.05 + 1/25 = 0.09. The approximation usedimplies that iM is greater than 0.09.Now use the general result that (1 + iR)(1 + e) = 1 + iM where iR is the real rate, e is the inflation rate and iM is themoney rate. This gives 1 + iR = 1.09/1.03 and hence iR ≈ 0.058. So iR is slightly more than 0.058.

If i = 0.058 then 0.5 + 1.030.25(1 + i)0.25 − 25.75i = 0.0282.If i = 0.059 then 0.5 + 1.030.25(1 + i)0.25 − 25.75i = 0.0027.If i = 0.0595 then 0.5 + 1.030.25(1 + i)0.25 − 25.75i = −0.010.Linear interpolation between 0.059 and 0.0595 gives (x−0.059)/(0.0595−0.059) = (f (x)−f (0.059)/(f (0.0595)−f (0.059)) and hence x = 0.059 + 0.0005(0− 0.0027)/(−0.010− 0.0027) = 0.0591 or 5.91%.

13. (i) Next expected dividend is d1; rate of dividend growth is g; let ν = 1/(1 + i). Then P =∑∞k=1 d1(1 + g)k−1νk =

d1ν/(1− (1 + g)ν) = d1/(i− g).(ii) Analyst I has g = 0; hence 750 = 35/i and i = 35/750 = 7/150 = 0.04667 or 4.667%.Analyst II has g = 0.1; hence 750 = 35/(i− 0.1) leading to i = 0.1 + 0.04667 = 0.14667 or 14.667%.


14. See answer to exercise 12. Let α = 1.015 and β = 1.04. The cash flow is as follows:Time (in years) 0 0.25 1.25 2.25 3.25 . . .Cash flow, unadjusted −125 5 5β 5β2 5β3 . . .Inflation factor 1 α0.25 α1.25 α2.25 α3.25 . . .

Hence we require:

125 =∞∑k=0

5βkν0.25+k

α0.25+k =5ν0.25

α0.25

∞∑k=0

βkνk

αk=

5ν0.25

1.0150.25

11− 1.04ν/1.015

and this leads to0 = 26ν + 1.0150.75ν0.25 − 25.375


0 = 0.625 + 1.0150.75(1 + i)0.75 − 25.375iFor an initial approximation, either of the methods in the answer to exercise 12 can be used. Using the first methodmeans we approximate (1 + i)0.75 by 1 + 0.75i. This leads to 0 = 0.625 + 1.0150.75(1 + 0.75i) − 25.375i and hencei =(0.625 + 1.0150.75

)/(25.375− 0.75× 1.0150.75

)= 0.0664. As in the answer to exercise 12, the value of i will

be larger than this approximation.If i = 0.0664, then 0.625 + 1.0150.75(1 + i)0.75 − 25.375i = 0.00128. If i = 0.067, then 0.625 + 1.0150.75(1 + i)0.75 −25.375i = −0.0135. Linear interpolation between 0.0664 and 0.067 gives (x− 0.0664)/(0.067− 0.0664) = (f (x)−f (0.0664)/(f (0.067)− f (0.0664)) and hence x = 0.0664 + 0.0006(0− 0.00128)/(−0.0135− 0.00128) = 0.0665 or6.65%.

15. (i) The cash flow sequence is:Time (in years) 0 4/12 10/12 16/12 . . .Cash flow −P d1 d1(1 + g)1/2 d1(1 + g) . . .

Hence

P =∞∑k=0

d1(1 + g)k/2

(1 + i)(4+6k)/12 =d1

(1 + i)4/12

∞∑k=0

(1 + g1 + i

)k/2

=d1(1 + i)1/6

(1 + i)1/2 − (1 + g)1/2

(ii) The cash flow sequence isTime (in years) 0 2/12 8/12 14/12 20/12 . . .Cash flow −£18 d1 d1(1 + g)1/2 d1(1 + g) d1(1 + g)3/2 . . .

Hence

18 =d1ν

2/12

1− ν1/2(1 + g)1/2 =0.5(1 + i)1/3

(1 + i)1/2 − 1.041/2

or18(1 + i)1/2 − 18× 1.041/2 = 0.5(1 + i)1/3

Initial approximation: replace (1 + i)1/2 by 1 + i/2 and (1 + i)1/3 by 1 + i/3. This leads to i = 0.096, rounding down.As in the answer to exercise 12, the value of i will be larger than this approximation.If i = 0.096, then 18(1 + i)1/2 − 18× 1.041/2 − 0.5(1 + i)1/3 = −0.0278.If i = 0.1, then 18(1 + i)1/2 − 18× 1.041/2 − 0.5(1 + i)1/3 = 0.0059.Hence answer lies between 9.6% and 10%, or 10% to nearest 1%.

16. (i) Are a share in the ownership of a company. Can be readily bought and sold. From point of view of investor:receive dividends and potential growth in share price. Size of dividends is not guaranteed. No fixed redemption time.Shareholders last in line to receive payment if company is liquidated. Shareholders can vote in AGM of company.

(ii) Denote the current price by P and the next dividend by d. Let α = 1.02, β = 1.03 and ν = 1/(1 + i) wherei = 0.05. Then the cash flow is as follows:

Time (in years) 0 0.5 1.5 2.5 3.5 . . .Cash flow −P d dβ dβ2 dβ3 . . .Inflation index 1 α0.5 α1.5 α2.5 α3.5 . . .

Hence

P = d∞∑k=0

βkνk+0.5

αk+0.5 =dν0.5

α0.5

11− βν/α

=dν0.5α0.5

α− βν=d(1 + i)0.5α0.5

α(1 + i)− β

and henced

P=

1.02× 1.05− 1.031.050.5 × 1.020.5 = 0.0396


17. (i) We need x where 10,000 = xa5 ,0.0625. Hence x = 10,000i/(1 − ν5) = 625/(1 − 1/1.06255) = 2390.13 or£2,390.13.

(ii) Investment A. Let a = 1.035 × 1.045 × 1.055 × 1.065 × 1.075. Then money interest earned is £10,000a.The January, 1993 value of this amount is £10,000a × 240/275.6. Hence (1 + iR)5 = a × 240/275.6 and soiR = (240.0a/275.6)1/5 − 1 = 0.0261 or 2.61%.Investment B. Redemption proceeds in January 1998 are 10,000 × (274.0/237.6) × 1.02755. Hence, if iM denotesthe effective money rate of return per annum, then (1 + iM )5 = (274.0/237.6)× 1.02755.Of course, we want the real rate of return, iR where (1 + iR)5 = (274.0/237.6) × 1.02755 × 240/275.6. HenceiR = (274.0× 240.0)/(237.6× 275.6)1/5 × 1.0275− 1 = 0.0284 or 2.84%.

(iii) Investment C. The cash flows are as follows (where x is given in part (i)):Date 1/93 1/94 1/95 1/96 1/97 1/98Cash flow −10,000 x x x x xInflation index 240 250 264.4 266.6 270.4 275.6

Hence the real rate of return, iR, satisfies

10,000 = x(

240250

νR +240

264.4ν2R +

240266.6

ν3R +

240270.4

ν4R +

240275.6

ν5R

)= 240xνR

(1

250+

νR264.4

+ν2R

266.6+

ν3R

270.4+

ν4R

275.6

)Question only asks which is the greatest real rate of return. So we need to decide if iR > 0.0284. SubstitutingiR = 0.0284 gives RHS ≈ 9967. Hence real yield from Investment C is less than 2.84% and the answer isInvestment B.

18. (i) Now g = 6/100 = 0.06 and t1 = 0.25. Hence i = 0.08 > (1− t1)g and so there is a capital gain. The cash flow isas follows:

Time (in years) 0 0.75 1.75 · · · 8.75 9.75Cash flow (before tax) −P 6 6 · · · 6 106Cash flow (after tax) −P 4.5 4.5 · · · 4.5 104.5− 0.25(100− P )

= 79.5 + 0.25P

Let x = 1/1.08. Hence:P = 4.5[x0.75 + x1.75 + · · · + x9.75] + (75 + 0.25P )x9.75

= 4.5x0.75[1 + x + · · · + x9] + (75 + 0.25P )x9.75

and hence

P (1− 0.25x9.75) = 4.5x0.75(

1− x10

1− x

)+ 75x9.75

and so P = 75.06.

(ii) Using (1 + iM ) = (1 + iR)(1 + q) gives 1.08 = 1.03(1 + q) and so q = 0.4854 or 4.854%.

19. Work in units of £100. Let α = 1.03. The cash flow is as follows:Date 0 0.5 1 1.5 · · · 25Payments, gross −99 4 4 4 · · · 4 + 110

The tax payments on the coupons will be 2 at time 1.25, 2 at time 2.25, . . . , and 2 at time 25.25. The CGT will be3.3 at time 25.25.

Let iM denote the money rate of return and let νM = 1/(1 + iM ). Then 99 = 8a(2)25 ,iM

+ 110ν25M − 2ν0.25

M a25 ,iM −3.3ν25.25

M =[8 ii(2) − 2ν0.25

M

]a25 ,iM + 110ν25

M − 3.3ν25.25M .

We need an approximate answer for iM . Note that the coupon is 6 per year after tax and the capital gain is 7.7 aftertax. Using method (d) in paragraph 4.3 gives iM ≈ (6 + 7.7/25)/99 = 0.0637 or 6.4%.If iM = 0.06, then

[8 ii(2) − 2ν0.25

M

]a25 ,iM + 110ν25

M − 3.3ν25.25M = 103.453.

If iM = 0.065, then[8 ii(2) − 2ν0.25

M

]a25 ,iM + 110ν25

M − 3.3ν25.25M = 97.241.

Using linear interpolation gives iM = 0.06 + 0.005× (99− 103.453)/(97.241− 103.453) = 0.0636.Now 1 + iM = (1 + q)(1 + iR) and hence iR = 1.0636/1.03− 1 = 0.032 or 3.2%.

(ii) If tax was collected later, then the real rate would increase.

20. See answer to exercise 12. Let α = 1.03 and β = 1.05. The cash flow is as follows:Time (in years) 0 0.667 1.667 2.667 3.667 . . .Cash flow, unadjusted −21.5 1.1 1.1β 1.1β2 1.1β3 . . .Inflation factor 1 α0.667 α1.667 α2.667 α3.667 . . .


Hence we require:

21.5 =∞∑k=0

1.1βkν0.667+k

α0.667+k =1.1ν0.667

α0.667

∞∑k=0

βkνk

αk=

1.1ν0.667

1.030.667

11− 1.05ν/1.03

and this leads to0 = 22.575ν + 1.1× 1.030.333ν0.667 − 22.145


0 = 0.43 + 1.1× 1.030.333(1 + i)0.333 − 22.145iFor an initial approximation, either of the methods in the answer to exercise 12 can be used. Using the first methodmeans we approximate (1 + i)0.333 by 1 + 0.333i. This leads to 0 = 0.43 + 1.1× 1.030.333(1 + 0.333i)− 22.145i andhence i =

(0.43 + 1.1× 1.030.333

)/(22.145− 0.333× 1.1× 1.030.333

)= 0.070, rounded down. As in the answer

to exercise 12, the value of i will be larger than this approximation.If i = 0.07, then 0.43 + 1.1 × 1.030.333(1 + i)0.333 − 22.145i = 0.0160. If i = 0.071, then 0.625 + 1.0150.75(1 +i)0.75 − 25.375i = −0.00575. Linear interpolation between 0.07 and 0.071 gives (x − 0.07)/(0.071 − 0.07) =(f (x) − f (0.07)/(f (0.071) − f (0.07)) and hence x = 0.07 + 0.001(0 − 0.016)/(−0.00575 − 0.016) = 0.0707 or7.07%.

21. The cash flows were as follows:Date 1/6/2000 1/12/2000 1/6/2001 1/12/2001 1/6/2002Payments, not indexed −94 1.5 1.5 1.5 1.5 + 100Payments, indexed −94 1.5× 102/100 1.5× 107/100 1.5× 111/100 (1.5 + 100)× 113/100

−94 1.53 1.605 1.665 1.695 + 113

(ii)(a) Before indexing, £94 grows to £113. But the purchase price of £94 is indexed to £94 × 118/102. Hence theCGT is £0.35× (113− 94× 118/102) = 1.489.(ii)(b) Let i denote the yield and let ν = 1/(1 + i). Then 94 = 0.75 × (1.53ν0.5 + 1.605ν + 1.665ν1.5 + 1.695ν2) +(113− 1.489)ν2. This leads to 376 = 4.59ν0.5 + 4.815ν + 4.995ν1.5 + 451.129ν2. Transforming into an equation in igives 376(1 + i)2 = 4.59(1 + i)1.5 + 4.815(1 + i) + 4.995(1 + i)0.5 + 451.129 or 376i2 + 747.185i− 79.944− 4.59(1 +i)1.5 − 4.995(1 + i)0.5 = 0.Using the usual approximation of (1 + i)0.5 ≈ 1 + 0.5i, etc gives 376i2 + 737.8i − 89.5 = 0. The usual formula forthe solution of a quadratic gives i = 0.11.If i = 0.11 then 376i2 + 747.185i− 79.944− 4.59(1 + i)1.5 − 4.995(1 + i)0.5 = −3.8344.If i = 0.12 then 376i2 + 747.185i− 79.944− 4.59(1 + i)1.5 − 4.995(1 + i)0.5 = 4.40588.Linear interpolation gives i = 0.11 + 0.01× (3.8344)/(4.40588 + 3.8344) = 0.1147 or 11.47%.

22. Work in units of £1,000. The cash flow is as follows:Date 0 1 2 3Payments, not indexed −25 10 10 10Retail Price Index 170.7 183.3 191.0 200.9Payments, indexed to time t = 0 −25 10× 170.7/183.3 10× 170.7/191.0 10× 170.7/200.9

(i)(a) Let iR denote the real rate of return and let νR = 1/(1 + iR). Then

25 =1707183.3

νR +1707191.0

ν2R +

1707200.9

ν3R or 25(1 + iR)3 − 1707

183.3(1 + iR)2 − 1707

191.0(1 + iR)− 1707

200.9= 0

Using the usual approximation of (1 + iR)3 ≈ 1 + 3iR, etc gives 1.7465 = 47.5iR and hence iR = 0.037.If i = 0.037 then 25(1 + iR)3 − 1707

183.3 (1 + iR)2 − 1707191.0 (1 + iR)− 1707

200.9 = 0.0998.If i = 0.03 then 25(1 + iR)3 − 1707

183.3 (1 + iR)2 − 1707191.0 (1 + iR)− 1707

200.9 = −0.2636.Using linear interpolation gives iR = 0.03 + 0.007× (0 + 0.2636)/(0.0998 + 0.2636) = 0.0351, or 3.51%.

(b) Let iM denote the money rate of return. Then 25 = 10a3 ,iM . Using tables shows that a3 ,0.095 = 2.508907 anda3 ,0.1 = 2.486852. Linear interpolation gives iM = 0.095+0.005×(2.5−2.508907)/(2.486852−2.508907) = 0.097or 9.7%.

(ii) Let e denote the average rate of inflation, then (1 + e)3 = 200.9/170.7 and hence e = 0.0558 or 5.58%. Theanswers to part (i) suggest 1 + e = (1 + iM )/(1 + iR) = 1.097/1.0351 = 1.0598 or 5.98%. They are not equal becausethe inflation index does not increase by a constant rate.

23. (i) See exercise 16. (ii) Let α = 1.06 and ν = 1/(1 + i) = 1/1.09. The cash flow is as follows:Date 1 2 3 4 5Dividends 25× 1.02 25× 1.02× 1.04 25× 1.02× 1.04α 25× 1.02× 1.04α2 · · ·

Hence

NPV = 25.5ν + 26.52ν2 + 26.52αν3 + 26.52α2ν4 + · · · = 25.51.09

+26.52ν2

1− αν=

25.5× 0.03 + 26.521.09× 0.03

= 834.40

(iii) The cash flow is as follows:


Date 0 1 2Cash flow −820 25× 1.02 25× 1.02× 1.04 + 900Inflation Index 1 1.03 1.03× 1.035

Hence

820 =25.5ν1.03

+926.3925ν2

1.06605or 874.161 = 26.3925ν + 926.52ν2

Solving the quadratic give ν = 0.957195 and hence i = 0.04472 or 4.47%.

24. (i) Let νM = 1/(1 + iM ) denote the money values. Then107 = 7.5a(2)

9 ,iM+ 100ν9

We need to find iM . Using method (d) in paragraph 4.3 gives iM ≈[7.5 + (100− 107)/9

]107 = 0.063.

Try iM = 0.065. Then RHS = 7.5× 1.015994× 6.656104 + 100× 0.567353 = 107.4545.Try iM = 0.07. Then RHS = 7.5× 1.017204× 6.515232 + 100× 0.543934 = 104.0983.Interpolation gives (iM − 0.065)/0.005 = (107− 107.4545)/(104.0983− 107.4545) and hence iM = 0.0657.Using the general result that (1 + iR)(1 + e) = 1 + iM where iR is the real rate, e is the inflation rate and iM is themoney rate, we get iR = 1.0657/1.025− 1 giving 3.97%.(ii) let α0 = 159.5/69.5 and α = 1.0251/2. The cash flow is as follows:

Date 9/10/97 8/4/98 8/10/98 · · · 8/4/06 9/10/06Time 0 1/2 1 · · · 81/2 9Cash flow 0 α0 α0α · · · α0α

16 α0α17

Hence if iM denotes the money rate of return with νM = 1/(1 + iM ), we have

NPV =18∑k=1

α0αk−1ν

k/2M + 100α0α

17ν9M

Now (1 + iR)(1 + e) = 1 + iM gives νR = (1 + e)νM = α2νM and so

NPV =18∑k=1

α0

ανk/2R + 100

α0

αν9R =

α0

α(2a(2)

9 ,iR+ 100ν9

R) = 2.2668(2a(2)9 ,iR

+ 100ν9R)

(iii) We want to find iR with 203 = 2.2668(2a(2)9 ,iR

+ 100ν9R) or

44.7767 = a(2)9 ,iR

+ 50ν9R =

iR

i(2)R

a9 ,iR + 50ν9R

Using method (d) in paragraph 4.3 gives i ≈[1 + (50− 44.7767)/9

]/44.7767 = 0.0353.

Using tables, iR = 0.03 gives RHS = 46.1649 and iR = 0.035 gives RHS = 44.3602Interpolating gives (x − 0.03)/(0.035 − 0.03) = (f (x) − f (0.03))/(f (0.035) − f (0.03)). Hence x = 0.03 +0.005(44.7767− 46.1649)/(44.3602− 46.1649) = 0.034 or 3.4%.

25. (i) Payments guaranteed by a government.Usually the values of the coupon and redemption payments are linked to some inflation index, with a time lag.Guaranteed real return if held to maturity—except for effect of time lag in indexation.The security of government payments tends to lead to a low volatility of return and a low expected return comparedto other investments.

(ii) Let α = 1.0250.5, the 6-month inflation factor and let x = 1/1.015, the 6-month discount factor. Cash flow is:

Date 1 July 03 1 Jan 04 1 July 04 1 Jan 05 · · · 1 Jan 09 1 July 09

Cash flow, not indexed −P 1 1 1 · · · 1 1 + 100Inflation index, 8 months earlier 113.2 113.8 113.8α 113.8α2 · · · 113.8α10 113.8α11

Cash flow, indexed, before tax −P 113.8110

113.8α110

113.8α2

110 · · · 113.8α10

110101×113.8

110

Cash flow, indexed, after tax −P 0.8×113.8110

0.8×113.8α110

0.8×113.8α2

110 · · · 0.8×113.8α10

110(0.8+100)×113.8α11

110

Inflation index 113.8α2/6 113.8α8/6 113.8α14/6 113.8α20/6 · · · 113.8α68/6 113.8α74/6

Equivalent Inflation index 1 α α2 α3 · · · α11 α12

Hence

P =0.8× 113.8x

110α+

0.8× 113.8αx2

110α2 +0.8× 113.8α2x3

110α3 + · · · + 0.8× 113.8α10x11

110α11 +100.8× 113.8α11x12

110α12

=0.8× 113.8x

110α+

0.8× 113.8x2

110α+

0.8× 113.8x3

110α+ · · · + 0.8× 113.8x11

110α+

100.8× 113.8x12

110α

=113.8x110α

[0.8(1 + x + x2 + · · · + x11) + 100x11]


=113.8x110α

[0.8

1− x12

1− x+ 100x11

]=

113.8110α

[0.8

1− x12

0.015+ 100x12

]= 94.383

26. (i) The coupon is £3/2 and this must be adjusted by the inflation index at 7/99 (eight months before 3/00). Hence theanswer is

32× 126.7

110.5= 1.72

(ii) Let x = 1.04 and α = 127.4/110.5 which is the inflation factor at 9/99 (used for 5/00, eight months later). Thecash flow is as follows:

Time (in years) 0 0.5 1 1.5 2 2.5Date 9/99 3/00 9/00 3/01 9/01 3/02Inflation factor 126.7

110.5 αx4/12 αx10/12 αx16/12 αx22/12

Cash flow −111 32

126.7110.5 1.5αx4/12 1.5αx10/12 1.5αx16/12 101.5αx22/12

Hence the yield, i satisfies νM = 1/(1 + i) where ν is given by

111 =32

(126.7110.5

ν1/2M + αx4/12νM + αx10/12ν

3/2M + αx16/12ν2

M + αx22/12ν5/2M

)+ 100αx22/12ν

5/2M

We want the real effective annual yield. Recall (1 + iR)(1 + e) = 1 + iM where iR is the real rate, e is the inflationrate and iM is the money rate.Hence νR = (1 + e)νM = xνM . Hence

111 =32

(126.7110.5

x−1/2ν1/2R + αx−8/12νR + αx−8/12ν

3/2R + αx−8/12ν2

R + αx−8/12ν5/2R

)+ 100αx−8/12ν

5/2R

Let γ = ν1/2R and define i1 by γ = 1/(1 + i1). Then

111 =32

(126.7110.5

x−1/2γ + αx−8/12γ2 + αx−8/12γ3 + αx−8/12γ4 + αx−8/12γ5)

+ 100αx−8/12γ5

=32

(126.7110.5

x−1/2 − αx−8/12)γ +

32αx−8/12a5 ,i1 + 100αx−8/12γ5

= 0.0017γ + 1.6848a5 ,i1 + 112.3186γ5

This must be solved numerically; so we need a rough starting value. Now an answer of 0.03 corresponds to i1 =1.031/2 − 1 = 0.014.

So try i1 = 0.015, then RHS = 112.320 (Use tables!)So try i1 = 0.020, then RHS = 109.673

Hence interpolation gives (x − 0.015)/(0.020 − 0.015) = (f (x) − f (0.015)/(f (0.020) − f (0.015)) and hence x =0.015 + 0.005(111− 112.320)/(109.673− 112.320) = 0.0175. Hence νR − 1 = 1/γ2 − 1 = (1 + i1)2 − 1 = 0.0353or 3.53%.

(iii) Increasing 110.5 means that γ must increase. Hence answer must decrease.

27. (i) Let α = 1.071/2 and β = 206/200. The cash flow is as follows:

Date 5/97 11/97 5/98 11/98 5/99 · · · 11/11 5/12Serial no. 0 1 2 3 4 · · · 29 30Inflation index 200 206 206α 206α2 206α3 · · · 206α28 206α29

Payments before adjustment −P 2 2 2 2 · · · 2 102Payments after adjustment −P 2β 2βα 2βα2 2βα3 · · · 2βα28 102βα29

Let iM denote the effective money rate of return and νM = 1/(1 + iM ).Let iR denote the effective real rate of return and νR = 1/(1 + iR).Hence (i + iR)(1.07) = i + iM and α2νM = νR. Also νR = 1.0152.

P =30∑k=1

2βαk−1νk/2M + 100βα29ν15

M =2βα

30∑k=1

νk/2R +

100βα

ν15R

P =2βα

a30 ,0.015 +100βα

11.01530 =

β

α

(2a30 ,0.015 +

1001.01530

)(b) Hence P = 111.53.

(ii) Now let γ = 1.051/2. The cash flow is now as follows (where P = 111.53):


Date 5/97 11/97 5/98 11/98 5/99 · · · 11/11 5/12Serial no. 0 1 2 3 4 · · · 29 30Inflation index 8 months ago 200 206 206γ 206γ2 206γ3 · · · 206γ28 206γ29

Payments before adjustment −P 2 2 2 2 · · · 2 102Payments after adjustment −P 2β 2βγ 2βγ2 2βγ3 · · · 2βγ28 102βγ29

The redemption money is 100βγ29 = 208.97.The RPI at 3/97 is 206 and hence 206γ1/3 at 5/97, the purchase date. The RPI at 5/12, 15 years later, is 206γ1/3γ30.Hence the purchase price revalued according to the RPI to the redemption date is 111.53 × γ30 = 231.86. Hencethere is no capital gains tax.Let iR denote the real effective yield per 6 months. Hence

111.53 =β

γ

(2a30 ,iR +

100(1 + iR)30

)=

1.03√1.05

(2a30 ,iR +

100(1 + iR)30

)and hence we need to solve

55.478 = a30 ,iR +50

(1 + iR)30

Using method (d) in paragraph 4.3 gives iR ≈[1 + (50− 55.478)/30

]/55.478 = 0.015.

if iR = 0.015, then RHS = 56.004. If iR = 0.02, then RHS = 50.0.Interpolating gives (x − 0.015)/(0.02 − 0.015) = (f (x) − f (0.015))/(f (0.02) − f (0.015)) and so x = 0.015 +0.005(55.478− 56.004)/(50.0− 56.004) = 0.0154 or 1.54%.

28. (i) Let α denote the rate of inflation and let ν = 1/1.06. Then the cash flow is as follows:

Time 0 1 2 3 . . .Cash flow in pence −260 12α 12α2 12α3 . . .

Hence 260 = 12(αν + α2ν2 + α3ν3 + · · ·) = 12αν/(1− αν). Hence α = (65× 1.06)/68 = 1.0132353 or 1.324%.(ii) Let α = 1.03 and ν = 1/(1 + i) where i is the rate of return. Then 260 = 12(αν + α2ν2 + · · · + α12ν12) + 500ν12.We want the real rate of return, i′, where 1.03(1 + i′) = 1 + i. So if x = 1/(1 + i′) = 1.03/(1 + i) = αν then260 = 12x(1 − x12)/(1 − x) + 500x12/1.0312. Let f (i′) = 260 − 12x(1 − x12)/(1 − x) − 500x12/1.0312. We havef (0.06) = −14.888284 and f (0.07) = 8.977237. Hence i′ ≈ 0.06 − 0.01f (0.06)/(f (0.07) − f (0.06)) = 0.066238or 6.6% per annum. (It is 6.61%.)

29. (i) The price is given by

P = 0.3× 1.051.09

+ 0.3× 1.052

1.092 + 0.3× 1.053

1.093 + · · · = 0.3× 1.051.09− 1.05

=638

= 7.875

or £7.875.(ii)(a) The return from the share is higher; hence P will be higher.(b) If profits increase with inflation and dividends are a constant percentage of profits, then the price may stay muchthe same because the investor will expect the same real rate of return.However, often the rate of inflation is just (incorrectly) added—hence 1.05 becomes 1.05 + I and 1.09 becomes1.09 + I where I is the increase in the inflation rate. Hence the new P is

0.3× (1.05 + I)0.04

which is greater than the old P .

(c) The risk the share will not actually give the returns specified is higher—hence the price P will be lower.

30. Let α = 1.03, β = 1.05, γ = 1.06 and ν = 1/1.08. (i) Required answer is3500

[αν + αβν2(1 + γν + γ2ν2 + · · ·)

]= 3500

[αν + αβν2/(1− γν)

]= 178581.01852 or £1,785.81.

(ii) Let i denote the required real rate of return and let ν = 1/(1 + i).Date: 1/1/2012 31/12/2012 31/12/2013 31/12/2014

Cash flow (in pounds): -1,720 35α 35αβ 35αβγ + 1,800Inflation Index: 110.0 112.3 113.2 113.8

Equation for ν is1,720110.0

=35α

112.3ν +

35αβ113.2

ν2 +35αβγ + 1,800

113.8ν3 or 172 = 3.53116652ν + 3.67824647ν2 + 177.86783963ν3.

So let f (i) = 3.53116652ν + 3.67824647ν2 + 177.86783963ν3− 172 where ν = 1/(1 + i). Now the dividend in 2012gives a real rate of return of (35α/1720)× (110/112.3) or 2%. Over the three years, there is also some capital gain.So try i = 0.03. Then f (0.03) = −2.330313; so it is less than 3%. Try 2.5%: f (0.025) = 0.114020. So the requiredvalue lies satisfies 0.025 < i < 0.03. Using linear interpolation gives i ≈ 0.025 + 0.005 × 0.114020/(2.330313 +0.114020) = 0.025233 or 2.52%.


31. Effective interest rate is i where 1 + i = 1.022 = 1.0404; hence i = 0.0404. Let ν = 1/(1 + i)1/12 = 1/1.04041/12. Letx be the required amount. Then we need 1.0520x to equal the cost of the annuity which is

10,000a(12)25 ,i =

25003

ν1− ν300

1− ν=

25003

1− 1/1.040425

1.04041/12 − 1= 158422.3048754

or £158,422.30 and hence x = 158,422.30/1.0520 = 59707.69872255 or £59,707.70.(ii) Now x grows to 1.0520x in 20 years. Adjusting for inflation, x grows to 1.0520x × (143/340) in 20 years. If jdenotes the annual effective real return, then (1 + j)20 = 1.0520x× (143/340) and hence j = 0.0055002 or 0.55%.(iii) Capital gain is x(1.0520− 1) and hence the tax is 0.25x(1.0520− 1) and the new payout is x

(3× 1.0520 + 1

)/4.

Let j′ denote the new real rate of return; then (1+j′)20 = (143/340)×(3× 1.0520 + 1

)/4. Hence j′ = −0.002977411

or -0.2977%. (iv) The tax is paid on the money capital gain. After deducting the tax, the remaining return isless than inflation which leads to a negative real return.

32. Adjusting for inflation, £95 grows to £100 × 220/222. Let i denote the required rate. Then 95(1 + i)91/365 =100× 220/222. Hence i =

((100× 220)/(95× 222)

)365/91 − 1 = 0.18463896 or 18.5%.

33. Investment A. After tax, the interest payments are £0.6 million per year. This is a net interest rate of 6%.Investment B. After 10 years, we have 1.110 million. The capital gains are 1.110 − 1. Amount returned after tax is1, 110−0.4(1 = .110−1) = 0.6×1.110+0.4. So if iB denotes the net rate of return, we have (1+iB)10 = 0.6×1.110+0.4which leads to iB = 0.06940531 or 6.94%.Investment C. Capital gains are 1.110−1.0410. Hence, amount returned is 0.6×1.110 +0.4×1.0410. So if iC denotesthe net rate of return, we have (1 + iC)10 = 0.6× 1.110 + 0.4× 1.0410 which leads to iC = 0.079469 or 7.95%.(ii) Gross rate of return is 10% for all 3 investments. Clearly, net rate for Investment C is greater than that forInvestment B because the capital gains are less and hence the tax is less. The net rate for Investment B is greaterthan that for Investment A because the tax is deferred until the end for Investment B and so greater compounding isachieved.

34. Let i be the effective money yield per annum and let ν = 1/(1+i); let α = 1.06. Then we have 175 = 6ν1/2 +6αν3/2 +6α2ν5/2 + · · · = 6ν1/2

[1 + αν + (αν)2 + · · ·

]= 6ν1/2/(1− αν) which gives the equation 185.5ν + 6ν1/2 − 175 = 0

or 175i − 6(1 + i)1/2 − 10.5 = 0. Approximating (1 + i)1/2 by 1 + i/2 gives i = 0.096. Trying i = 0.095 givesLHS = −0.1535349 and trying i = 0.096 gives LHS = 0.0185989. Linear interpolation gives i = 0.09589 or9.589%. Hence if iR denotes the real rate, then 1 + iR = 1.09589/1.04 leading to iR = 0.0537 or 5.37%.

35. (i) See §4.1 on page 73. (ii) If dividends are assumed to grow by a constant proportion, g, each year, then the cashflow is:

Time 0 1 2 3 . . .Cash flow, before tax −P d d(1 + g) d(1 + g)2 . . .Cash flow, after tax −P (1− t)d (1− t)d(1 + g) (1− t)d(1 + g)2 . . .

and so the rate of return r of the share given by:

P =∞∑k=1

(1− t)d(1 + g)k−1

(1 + r)k=

(1− t)dr − g

(iii) So d = 6, g = 0.01, r = 0.06 and t = 0.2. Hence P = 96p. (iv) The required return, r would increase; hence Pwould decrease. (v) Cost on 1 August 2014 is £960. On 1 August 2015, receives dividend income £60 gross or £48net and sells shares for £1200 with CGT of£60. Hence £960 grows to £1,188. Let r denote the net real rate of return;then 960(1 + r)/123 = 1,188/126. Hence

1 + r =123× 1,188126× 960

=41× 3314× 80

= 1.208036

or 20.80%.


1. Let f1,2 denote the required value. Then 1.08(1 + f1,2)2 = 1.0953. Hence f1,2 =√

1.0953/1.08 − 1 = 0.10258 or10.26%.

2. Let f1,2 denote the answer. Then 1.045(1 + f1,2)2 = 1.0553. Hence f1,2 =√

1.0553/1.045 − 1 = 0.060036 or6.0036%.

3. For the par yield, we want the coupon rate, r which ensures that the bond price equals the face value. Hence wewant:

Time 0 1 2 3Cash Flow −1 r r 1 + r

Hence

1 =r

1 + y1+

r

(1 + y2)2 +1 + r

(1 + y3)2


= r(

11.06

+1

1.06× 1.065+

11.06× 1.065× 1.07

)+

11.06× 1.065× 1.07

and hence

r =1.06× 1.065× 1.07− 11.065× 1.07 + 1.07 + 1

= 0.06478

or 6.478%.

4. Let r denote the required answer. Then

1 = r(

11.06

+1

1.0652 +1

1.06× 1.0662

)+

11.06× 1.0662

and hence

r =1.06× 1.0662 − 1

1.0662 + 1.06× 1.0662/1.0652 + 1= 0.06395 or 6.395%.

5. (i) Ft,1 = ln(1 + ft,1) = ln 1.04 = 0.39221. (ii) Ft = limk→0 Ft,k = limk→0 ln(1 + ft,k) where Ft,k is the forwardforce of interest of an investment that starts at time t and lasts for k years.

6. Let yk denote the k-year spot rate of interest. Then “the two-year par yield at time t = 0 is 4.15%” implies

1 = 0.0415[

11 + y1

+1

(1 + y2)2

]+

1(1 + y2)2

The other condition implies

105.4 =8

1 + y1+

106(1 + y2)2

Thus if a = 1/(1 + y1) and b = 1/(1 + y2)2 we have the 2 simultaneous equations:1 = 0.0415a + 1.0415b

105.4 = 8a + 106bThis leads to (106 − 8 × 1.0415/0.0415)b = 105.4 − 8/0.0415 and hence b = 0.92192. Similarly, (106 ×0.0415/1.0415− 8)a = 106/1.0415− 105.4 and hence a = 0.9596.So y1 = 1/a− 1 = 0.0421, or 4.21%, and y2 =

√1/b− 1 = 0.0415, or 4.15%.

7. We are given that yn = 0.0225 + 0.0025n for n = 1, 2, . . . , 5.(a) We want f3,2. But (1 + y5)5 = (1 + y3)3(1 + f3,2)2. Hence (1 + f3,2)2 = 1.0355/1.033. Hence f3,2 = 0.04255 or4.255%. (b) Let iP denote the 4-year par yield. Then

1 = iP

[1

1 + y1+

1(1 + y2)2 +

1(1 + y3)3 +

1(1 + y4)4

]+

1(1 + y4)4

and hence 1 = 3.717853iP + 0.8799 and hence iP = 0.0323 or 3.23%. (c) Par yield bond:Time 0 1 2 3 4Cash flow for par yield bond −100 3.23 3.23 3.23 100 + 3.23Cash flow for other bond, bond B −PB 3.5 3.5 3.5 100 + 3.5

The gross redemption yield of the par yield bond is clearly the same as the par yield; this is iP = 0.0323. Writingthis as an equation:

100 =3.23

1 + iP+

3.23(1 + iP )2 +

3.23(1 + iP )3 +

103.23(1 + iP )4 =

3.231 + y1

+3.23

(1 + y2)2 +3.23

(1 + y3)3 +103.23

(1 + y4)4

Let iB denote the gross redemption yield and PB denote the price of the other bond. Then

PB =3.5

1 + iB+

3.5(1 + iB)2 +

3.5(1 + iB)3 +

103.5(1 + iB)4 =

3.51 + y1

+3.5

(1 + y2)2 +3.5

(1 + y3)3 +103.5

(1 + y4)4

Now [3.5

1 + y1+

3.5(1 + y2)2 +

3.5(1 + y3)3 +

103.5(1 + y4)4

]−[

3.51 + iP

+3.5

(1 + iP )2 +3.5

(1 + iP )3 +103.5

(1 + iP )4

]=

3.53.23

[100

(1 + iP )4 −100

(1 + y4)4

]+ 100

[1

(1 + y4)4 −1

(1 + iP )4

]=[

3.5× 1003.23

− 100] [

1(1 + iP )4 −

1(1 + y4)4

]> 0

Hence

PB >

[3.5

1 + iP+

3.5(1 + iP )2 +

3.5(1 + iP )3 +

103.5(1 + iP )4

]and hence iB < iP .


8. (i) The n-year spot rate of interest is the per annum yield to maturity of a zero coupon bond with n years to maturity.(ii) Let f3,2 denote the answer. Then 1.0755 = 1.063(1 + f3,2)2 gives f3,2 =

√1.0755/1.063− 1 = 0.09790 or 9.79%.

(iii) Let r denote the 6-year par yield. Then

1 = r(

11.04

+1

1.052 +1

1.063 +1

1.074 +1

1.0755 +1

1.086

)+

11.086

and hence

r =1.086 − 1

1.086(1/1.04 + 1/1.052 + 1/1.063 + 1/1.074 + 1/1.0755) + 1= 0.07708

or 7.708%.

9. (i)(a)We have A(8) = exp(∫ 8

0 δ(t) dt)

= exp(0.32 + 0.032) = 1.421909. (b) We have A(9) = exp(∫ 9

0 δ(t) dt)

=

exp(0.36 + 0.0405) = exp(0.4005) = 1.492571. (c) We need exp(∫ 9

8 δ(t) dt) = A(9)A(8) = exp(0.4005)/ exp(0.352) =

1.049695. (ii)(a) We need y8 with (1 + y8)8 = A(8) = exp(0.352) and hence y8 = exp(0.044) − 1 = 0.044982or 4.4982%. (b) We need y9 with (1 + y9)9 = A(9) = exp(0.4005) and hence y9 = exp(0.0445) − 1 = 0.045505or 4.5505%. (c) We want f8,1 where (1 + f8,1)(1 + y8)8 = (1 + y9)9 and hence f8,1 = exp(0.4005)/ exp(0.352) =exp(0.0485) = 1.0496954. Hence f8,1 = 0.0496954 or 4.96954%.

10. (i) Let ν = 1/(1 + i) where i is the gross redemption yield p.a. Then 98 = 7ν + 7ν2 + 112ν3.The usual approximation is i ≈ (7 + 7/3)/98 = 0.0952. If i = 0.09 then RHS = 98.798. If i = 0.1 thenRHS = 96.296.Using linear interpolation, (0.09− α)/0.798 = −0.01/2.502. Hence α = 0.09319 or 9.319%.(ii) For the 1-year bond, ν1 = 98/112 = 7/8 and y1 = 1/7. Hence 1-year spot rate is y1 = 1/7 or 14.286%.Using the 2-year bond gives 98 = 7ν1 + 112ν2

2 or 16ν22 = 14 − ν1 which gives ν2 = 0.9057 and y2 = 0.10410 or

10.41%.Using the 3-year bond gives 98 = 7ν1 + 7ν2

2 + 112ν33 or 16ν3

3 = 14 − ν1 − ν22 = 15(14 − ν1)/16 = which gives

ν3 = 0.91619 and y3 = 0.09148 or 9.148%.

11. Let ft,r denote the forward interest rate p.a. for an agreement starting at time t for r years. We are given (1 + f0,2)2 =1118/1000; (1 + f1,2)2 = 1140/1000; 1 + f1,1 = 1058/1000.(i) (a) (1 + f0,2)2 = (1 + f0,1)(1 + f1,1). Hence 1 + f0,1 = 1118/1058 and f0,1 = 0.0567. (b) (1 + f0,2)2 = 1118/1000.Hence f0,2 = 0.057355. (c) (1 + f0,3)3 = (1 + f0,1)(1 + f1,2)2 = (1118/1058)× (1140/1000). Hence f0,3 = 0.06403.(ii) Three year par yield is coupon rate r which causes current bond price to be equal to its face value, assuming it isredeemed at par.

Time 0 1 2 3Cash Flow, c −100 100r 100r 100 + 100r

Hence

1 =r

1 + f0,1+

r

(1 + f0,2)2 +1 + r

(1 + f0,3)3 and so 1 = r[

10581118

+10001118

+1058× 10001118× 1140

]+

1058× 10001118× 1140

Hence r = 0.0636.

12. (i) If f1,1 denotes the implied one-year forward rate applicable at time 1, then (1 + y1)(1 + f1,1) = (1 + y2)2 and hence1.041(1 + f1,1) = 1.0422 and f1,1 = 0.0430 or 4.30%.If f2,1 denotes the implied one-year forward rate applicable at time 2, then (1 + y2)2(1 + f2,1) = (1 + y3)3 and hence1.0422(1 + f2,1) = 1.0433 and f2,1 = 0.0450 or 4.50%.

(ii) The cash flow is as follows:Time 0 1 2 3Cash flow −P 3 3 113

Hence

P =3

1 + y1+

3(1 + y2)2 +

113(1 + y3)3 = 105.24

(b) The 2-year par yield is the coupon rate that causes the bond price to be equal to its face value, assuming the bondis redeemed at par.Let yp2 denote the 2-year par yield. Then the cash flow is as follows:

Time 0 1 2Cash flow −100 100yp2 100yp2 + 100

This gives

1 = yp2

(1

1 + y1+

1(1 + y2)2

)+

1(1 + y2)2 = yp2

(1

1.041+

11.0422

)+

11.0422

and hence yp2 = (1.0422 − 1)/(1.0422/1.041 + 1) = 0.04198 or 4.198%.


13. (i)(a)

P =5

1.04+

51.04× 1.045

+105

1.04× 1.045× 1.048= 101.60

(b) The gross redemption yield is the value for i which satisfies 101.6 = 5a3 ,i + 100ν3.The usual approximation (method (d) in paragraph 4.3) gives i ≈ (5− 1.6/3)/101.6 = 0.044.If i = 0.045, then right hand side = 101.375. If i = 0.04, then right hand side = 102.785.Hence i− 0.45 = (101.6− 101.375)(0.04− 0.045)/(102.785− 101.375). Hence i = 0.0442 or 4.42%.(ii) The gross redemption yield is a weighted average of the forward rates. A higher coupon means more weight onthe coupon at times 1 and 2 when the rate is lower. Hence the yield will be lower.

14. (i) For the 3-year bond:Time 0 1 2 3Cash flow −96 6 6 106

Hence 96 = 6a3 ,i + 100ν3. The usual approximation (method (d) in paragraph 4.3) gives i ≈ (6 + 4/3)/96 = 0.076.At 7.5%,RHS = 6×2.600526+80.4961 = 96.099. At 8%,RHS = 6×2.577097+79.3832 = 94.846. Interpolatinggives (x− 0.075)/(0.075− 0.07) = (96− 96.099)/(96.099− 97.375) and hence x = 0.0754 or 7.54%.(ii) One year bond: 96 = ν(1)106. Hence ν(1) = 96/106 = 0.90566.Two year bond: 96 = 6ν(1) + 106ν(2). Hence ν(2) = (96.100)/1062 = 0.85440.Three year bond: 96 = 6ν(1) + 6ν(2) + 106ν3. Gives ν(3) = 96.1002/1063 = 0.80603.(iii) 1 + f0,1 = 1 + y1 = 106/96. Hence f0,1 = 10/96 = 0.1042 or 10.42%.Using (1 + y2)2 = (1 + y1)(1 + f1,2) gives 1062/(96.100) = (106/96)(1 + f1,2) and so f1,2 = 6/100 = 0.06 or 6%.Using (1 + y3)3 = (1 + y2)2(1 + f2,3) gives 1063/(96.1002) = (1062/(96.100))(1 + f2,3) and so f2,3 = 6/100 = 0.06 or6%.

15. (i)(a) One-year spot rate: 1+y1 = 100/97 and so y1 = 3/97 = 0.0309278 or 3.0928%. Two-year spot rate: (1+y2)2 =100/93 and so y2 = 0.0369517 or 3.6952%. Three-year spot rate: (1 + y3)3 = 100/88 and so y3 = 0.0435320 or4.3532%. Four-year spot rate: (1 + y4)4 = 100/83 and so y4 = 0.0476844 or 4.7684%.(b) Now

P =4× 97

100+

4× 93100

+4× 88

100+

104× 83100

= 97.44

Let i denote the rate of return. Then 97.44 = 4a4 ,i + 100ν4. The usual approximation (method (d) in paragraph 4.3)gives i ≈ (4 + (100− 97.44)/4)/97.44 = 0.048.Try i = 0.05, then 4a4 ,i + 100ν4 − 97.44 = −0.985996. Try i = 0.045, then 4a4 ,i + 100ν4 − 97.44 = 0.766204.Using linear interpolation gives (i−0.045)/0.005 = (0−0.766204)/(−0.985996−0.766204) leading to i = 0.04719or 4.72%.(ii) The rate of return from the bond is a weighted average of the rates for 1 year, 2 years, 3 years and 4 years. The1 year, 2 year and 3 year rates are all less than the 4 year spot rate.(iii) The liquidity preference theory asserts that investors prefer short term investments over long term ones. This isconsistent with the answers in part (i)(a): the rates increase as the term increases.

16. The gross redemption yield from a one year bond with a 6% annual coupon is 6% per annum effective:Time 0 1Cash flow −P 106

Hence P = 106/1.06 = 100. Hence i1 = f0,1 = 0.06.

The gross redemption yield from a 2-year bond with a 6% annual coupon is 6.3% per annum effective:Time 0 1 2Cash flow −P 6 106

Hence P = 6/1.063 + 106/1.0632 = 6/(1 + f0,1) + 106/(1 + f0,2)2 = 6/1.06 + 106/(1 + f0,2)2. Hence i2 = f0,2 =0.0630905.

The gross redemption yield from a 3-year bond with a 6% annual coupon is 6.6% per annum effective. HenceP = 6/1.066 + 6/1.0662 + 106/1.0663 = 6/(1 + f0,1) + 6/(1 + f0,2)2 + 106/(1 + f0,3)2

= 6/1.06 + 6/1.06309052 + 106/(1 + f0,3)3

Hence i3 = f0,3 = 0.066247 or 6.6247%.Summary: i1 = 0.06; i2 = 0.0630905 and i3 = 0.066247.(b) We already know that f0,1 = 0.06. Now (1 + f0,1)(1 + f1,1) = (1 + i2)2; hence f1,1 = 0.06619.Now (1 + f0,1)(1 + f1,1)(1 + f2,1) = (1 + i3)3; hence f2,1 = 0.07259.Summary: f0,1 = 0.06; f1,1 = 0.06619 and f2,1 = 0.07259.(ii) Spot rates are geometric averages of forward rates: for example, 1 + i2 =

√(1 + f0,1)(1 + f1,1), etc. If forward

rates are increasing, then the spot rates will increase more slowly—because the geometric average depends on theinitial lower rates also.


17. (i)(a) Bond yields are determined by expectations of future interest rates. If interest rates are expected to fall, thenlong term bonds will become more attractive, and conversely. (b) Liquidity preference: short term investments arepreferred over longer term investments—hence longer term bonds should offer higher returns. Market segmenta-tion: rates reflect supply and demand for bonds of different lengths.(ii)(a) We are given y1 = 0.1, f1,1 = 0.09, f2,1 = 0.08 and f3,1 = 0.07.Gross redemption yield from a 1-year zero coupon bond: i = 0.1.Gross redemption yield from a 3-year zero coupon bond: i = (1.1× 1.09× 1.08)1/3 − 1 = 0.08997.Gross redemption yield from a 5-year zero coupon bond: i = (1.1× 1.09× 1.08× 1.072)1/5 − 1 = 0.08194.Gross redemption yield from a 10-year zero coupon bond: i = (1.1× 1.09× 1.08× 1.077)1/10 − 1 = 0.07595.(b) The values decrease.(c) The gross redemption yield of a 5-year coupon-paying bond can be regarded as a weighted average of the grossredemption yields for zero-coupon bonds of shorter terms—see part (c) of example(1.2a). And the early couponsattract higher rates. Hence the gross redemption yields of coupon-paying bonds will decrease with time more slowlythan the gross redemption yields of zero coupon bonds.

18. We are given that yn = 0.08− 0.04e−0.1n. (i) P = 1/(1 + y9)9 = 1/(1.08− 0.04e−0.9)9 = 0.57344375. (ii) Wehave (1 + f7,4)4(1 + y7)7 = (1 + y11)11 and hence (1 + f7,4)4 = (1.08 − 0.04e−1.1)11/(1.08 − 0.04e−0.7)7 leading tof7,4 = 0.078243 or 7.8243%. (iii) We need r with

1 =r

1 + y1+

r

(1 + y2)2 +1 + r

(1 + y3)3

leading to

r =[

1− 1(1 + y3)3

]/[1

1 + y1+

1(1 + y2)2 +

1(1 + y3)3

]=[

1− 1(1.08− 0.04e−0.3)3

]/[1

1.08− 0.04e−0.1 +1

(1.08− 0.04e−0.2)2 +1

(1.08− 0.04e−0.3)3

]= 0.050157

or 5.0157%.19. (i) Now (1 + f2,2)2 = (1 + f2,1)(1 + f3,1). This implies 1.052 = 1.045(1 + f3,1) and hence f3,1 = 1.052/1.045 − 1 =

0.0550239 or 5.5024%. (ii) One year: y1 = 0.04 or 4%. Two year: y2 =√

1.04× 1.0425 − 1 = 0.04124925 or4.125%. Three year: y3 = (1.04× 1.0425× 1.045)1/3 − 1 = 0.042498 or 4.25%. Four year: y4 = (1.04× 1.0425×1.052)1/4 − 1 = 0.0456155 or 4.56%. (iii) Cash flow from bond is as follows:

Time 0 1 2 3 4Cash flow −P 3 3 3 103

Hence

P =3

1.04+

31.04× 1.0425

+3

(1 + y3)3 +103

(1 + y4)4 = 94.46813

Let i denote gross redemption yield. Hence

94.46813 =3

1 + i+

3(1 + i)2 +

3(1 + i)3 +

103(1 + i)4 = 3a4 ,i +

100(1 + i)4

Let f (i) = 3a4 ,i + 100/(1 + i)4 − 94.46813. Then f (0.045) = 3 × 3.587526 + 83.8561 − 94.46813 = 0.150548and f (0.05) = 3 × 3.545951 + 82.2702 − 94.46813 = −1.560077. Interpolating gives (i − 0.045)/(−0.150548) =0.005/(−1.560077− 0.150548) leading to i = 0.04544 or 4.544%. (iv) We have

31 + i

+3

(1 + i)2 +3

(1 + i)3 +103

(1 + i)4

=3

1 + f0,1+

3(1 + f0,1)(1 + f1,1)

+3

(1 + f0,1)(1 + f1,1)(1 + f2,1)+

103(1 + f0,1)(1 + f1,1)(1 + f2,1)(1 + f3,1)

It follows thatminf0,1, f1,1, f2,1, f3,1 < i < maxf0,1, f1,1, f2,1, f3,1

The gross redemption yield, i, is a complicated average of the 1-year forward rates.

20. (i) The spot rates are i1 = a − b and i2 = a − 2b. Also f01 = 0.061 and f11 = 0.065. Hence a − b = 0.061and (1 + f01)(1 + f11) = (1 + a − 2b)2. Hence b = 1.061 −

√1.061× 1.065 = −0.00199812 and a = 1.122 −√

1.061× 1.065 = 0.05900188.(ii) First we need i4, the 4-year spot rate. Using the fact that the 4-year par yield is 0.07, we get

1 = 0.07(

11 + a− b

+1

(1 + a− 2b)2 +1

(1 + a− 3b)3

)+

1.07(1 + i4)4

Hence i4 = 0.07071303. Then

P = 0.05(

11 + a− b

+1

(1 + a− 2b)2 +1

(1 + a− 3b)3

)+

1.08(1 + i4)4 = 0.954501698798

or £0.9545 per £1 nominal.


21. (i) Let i denote the gross redemption yield and let ν = 1/(1 + i). Then 103 = 6ν + 6ν2 + 111ν3. Spreading thecapital gain of £2 equally over the 3 years, suggests i ≈ 6.7/103 = 0.065. So try i = 0.065. Then 6ν + 6ν2 +111ν3 − 103 = −0.1849923. Try i = 0.06; then 6ν + 6ν2 + 111ν3 − 103 = 1.1980964. Linear interpolation givesi = 0.06 + 0.005× 1.1980964/(1.1980964 + 0.1849923) = 0.064331 or 6.43%.(ii) The 1 year spot rate, i1 is given by 111 = (1 + i1)103 and hence i1 = 0.0776699 or 7.767%. The 2 year spot rate,i2, is given by 103 = 6/(1 + i1) + 111/(1 + i2)2 and hence i2 = 111/

√103× 111− 618− 1 = 0.067357 or 6.736%.

The 3-year spot rate, i3 is given by 103 = 6/(1 + i1) + 6/(1 + i2)2 + 111/(1 + i3)3. Hence i3 = 0.0639414 or 6.394%.(iii) To answer to 3 decimal places, we need more than 3 decimal places for the spot rates. Now f01 = i1 = 0.07767,f02 = i2 = 0.06736 and f03 = i3 = 0.06394.Now (1 + i1)(1 + f12) = (1 + i2)2 and hence f12 = 0.05714286 or 5.714%.Now (1 + i2)2(1 + f23) = (1 + i3)3 and hence f23 = 0.05714286 or 5.714%.Now (1 + i1)(1 + f13)2 = (1 + i3)3 and hence f13 = 0.05714286 or or 5.714%.In fact f12 = f23 = f13.

22. (i) Let yn = 0.06− 0.02e−0.1n. Then the price per £100 of the bond is P = 3/(1 + y1) + 3/(1 + y2)2 + 103/(1 + y3)3 =95.844762. Let i denote the gross redemption yield. Then P = 3/(1 + i) + 3/(1 + i)2 + 103/(1 + i)3. Now the capitalgain is approximately £4.16 over 3 years. This suggests a yield of 3 + 1.4 = 4.4% approximately. So try i = 0.044.Let f (i) = P − 3ν − 3ν2 − 103ν3 where ν = 1/(1 + i). Then f (0.044) = −0.299419; f (0.046) = 0.234823. Usinglinear interpolation gives i− 0.044/(−f (0.044)) = 0.002/(f (0.046)− f (0.044)) and hence i ≈ 0.0451 or 4.5%.(ii) The 4-year par yield, r satisfies 1 = r/(1 + y1) + r/(1 + y2)2 + r/(1 + y3)3 + (1 + r)/(1 + y4)4. Hence r =1− 1/(1 + y4)4)/(1/(1 + y1) + 1/(1 + y2)2 + 1/(1 + y3)3 + 1/(1 + y4)4) = 0.046424 or 4.64%.

23. (i) and (ii)(a) See answer to question 9 on page 174.(ii) Let x denote the forward rate of interest per annum convertible monthly in the second month. Then 1.01(1 +x/12) = 1.022. Hence x = 0.3611881 or 36.12%.

24. (i) Let i denote the gross redemption yield (which is the same as the yield to maturity or the internal rate of returnwhen taxation is ignored). Let ν = 1/(1 + i). Then 97 = 6ν + 6ν2 + 109ν3 or 97(1 + i)3−6(1 + i)2−6(1 + i)−109 = 0or f (i) = 97i3 + 285i2 + 273i− 24 = 0. The usual approximation gives i = 0.06 + 6/(3× 100) = 0.08.If i = 0.08, then f (i) = −0.286336 and if i = 0.085 then f (i) = 1.323695. Linear interpolation gives i = 0.08 +0.286336× 0.005/(1.323695 + 0.286336) = 0.080889 or 8.09%.(ii) Let yt denote the t-year spot rate of interest. Then 97(1 + y1) = 109. Hence y1 = 12/97 = 0.12371 or 12,37%.For y2 we have

97 =6

1 + y1+

109(1 + y2)2 =

6× 97109

+109

(1 + y2)2

Hence y2 = 0.09049 or 9.049%.

25. Now∫ t

0 δ(s) ds = 0.05t + 0.001t2.(a) exp(

∫ 70 δ(s) ds) = exp(0.35 + 0.049) = exp(0.399) = 1.490334 or 1.4903. (b) exp(0.30 + 0.036) = exp(0.336) =

1.399339 or 1.3993 (c) exp(0.399)/ exp(0.336) = exp(0.063) = 1.065027 or 1.0650.(ii)(a) If yt denotes the required spot rate of interest, then (1 + yt)7 = exp(0.399). Hence yt = exp(0.399/7) − 1 =0.058656 or 5.866%. (b) exp(0.336/6) − 1 = 0.057598 or 5.760%. (c) We want f6,1 where exp(0.399) =exp(0.336)(1 + f6,1). Hence f6,1 = exp(0.063)− 1 = 0.065027 or 6.503%.(iii) The formula for δ(t) shows that the interest rate is increasing with t. Hence the interest rate in the seventh year,which is the answer to part (ii)(c), will be higher that the average interest rate over the first seven years, which is theanswer to part (ii)(a).(iv) Now ρ(t) = 30e−0.01t+0.001t2

for 3 < t < 10 and ν(t) = e−0.05t−0.001t2. Hence the present value is∫ 10

3ρ(t)ν(t) dt = 30

∫ 10

3e−0.06t dt = 30

e−0.06t

−0.06

∣∣∣∣10

3= 500(e−0.18 − e−0.6) = 143.229287659 or 143.229.

26. (i) Spot rates of interest are y1/2 = 0.01 for t = 1/2; y3/2 = 0.03 for t = 3/2; y5/2 = 0.05 for t = 5/2; y7/2 = 0.07 fort = 7/2; and y9/2 = 0.09 for t = 9/2. Hence NPV = 1/1.011/2 + 1/1.033/2 + 1/1.055/2 + 1/1.077/2 + 2/1.099/2 =4.98307921162 or £4,983,079.(ii) Liquidity preference: investors prefer short term investments and hence longer term bonds must offer higheryields as an inducement. Expectations theory: investors expect interest rates to rise. Market segmentation theory:there is less demand for long-term bonds.(iii) Now (1 + y9/2)9/2 = (1 + y7/2)7/2(1 + f7/2) or 1.099/2 = 1.077/2(1 + f7/2). Hence f7/2 = 0.1629901 or 16.3%.

27. (i) Let ν = 1/(1 + i). Then 101 = 3ν + 3ν2 + 103ν3. Let f (i) = 103ν3 + 3ν2 + 3ν − 101. The value of i isclearly less than 3%. Now f (0.03) = −1 and f (0.025) = 0.428012. Linear interpolation gives i = 0.025 + 0.005 ∗0.428012/1.428012 = 0.0264986 or 2.65%. (It is actually 2.64885%.)(ii) One-year bond: 101(1 + i1) = 103. Hence i1 = 2/101 = 0.01980198 or 1.980%. Two-year bond: 101 =3/(1 + i1) + 103/(1 + i2)2; hence 103/(1 + i2)2 = (101× 103− 303)/103. Hence i2 = 0.0248883 or 2.489%. Three-year bond: 101 = 3/(1 + i1) + 3/(1 + i2)2 + 103/(1 + i3)3 = 303/103 + 3(101 × 103 − 303)/1032 + 103/(1 + i3)3;hence i3 = 0.026589379 or 2.659%. (iii) Now (1 + i2)2(1 + f2,1) = (1 + i3)3. Hence f2,1 = 0.03 exactly or 3%. (iv)See notes: expectations theory, liquidity preference, market segmentation with explanation.



1.Time 0 0.5 1 · · · 9.5 10Cash flow −P 5 5 · · · 5 105

Let j = 0.05 and x = 1/(1 + j) = 20/21. Also let 1 + i = (1 + j)2. Now P = 5a20 ,j + 100x20 = 100. HencedM (i) =

∑nk=1 tkctkν

tk/P where the numerator is 2.5(Ia)20 ,j + 100× 10x20. Using the general result that (Ia)n =(an −nνn+1)/(1− ν) implies the numerator equals 2.5× (a20 ,j − 20x21)× 21 + 1000x20 = 654.266. Hence answeris 6.543 years.Or from first principles:

dM (i) =1

100[2.5(x + 2x2 + 3x3 + · · · + 20x20) + 1000x20]

=1

100

[2.5(x(1− x20)(1− x)2 −

20x21

1− x

)+ 1000x20

]=

1100

[2.5(20× 21(1− x20)− 20× 21x21) + 1000x20] = 10.5

(1− x20)

2. Now

dM (i) =∑nk=1 tkctkν

tk∑nk=1 ctkν

tkand

dν

di= −ν2

Henced

didM (i) = −ν2

∑nk=1 ctkν

tk∑nk=1 t

2kctkν

tk − (∑nk=1 tkctkν

tk )2

(∑nk=1 ctkν

tk )2

= −ν2

[∑nk=1 t

2kctkν

tk∑nk=1 ctkν

tk−(∑n

k=1 tkctkνtk∑nk=1 ctkν

tk

)2]

= −ν2σ2

and this quantity is clearly negative.

3. First the price:

P = 10∫ 20

0νt dt = 10

νt

ln ν

∣∣∣∣20

0=

10(ν20 − 1)ln ν

Hence

dM (i) =10P

∫ 20

0tνt dt =

10P

[tνt

ln ν

∣∣∣∣20

0−∫ 20

0

νt

ln νdt

]=

10P

[20ν20

ln ν− νt

(ln ν)2

∣∣∣∣20

0

]

=10

P (ln ν)2

[20ν20 ln ν − ν20 + 1

]=

1− ν20 − 20ν20 ln 1.04(1− ν20) ln 1.04

= 8.70586

or 8.706 years.

4. (a) We must have (1) VA(i0) = VL(i0), the net present values are the same; (2) dA(i0) = dL(i0), the effective durationsare the same; and (3) cA(i0) > cL(i0), the convexity of the assets is greater than the convexity of the liabilities.(b) We have dM (i) = (1 + i)d(i) where dM is the duration and d is the volatility.(c) Now P (i) = ν + ν2 + · · · = ν/(1− ν) = 1/i. Hence

dP (i)di

= − 1i2

and so − 1P

dP (i)di

=1i

5. Let ν = 1/1.08. We need to checkn∑k=1

atkνtk =

n∑k=1

ltkνtk

n∑k=1

tkatkνtk =

n∑k=1

tkltkν)tkn∑k=1

t2katkνtk >

n∑k=1

t2kltkνtk

First equality: 12.425ν12 + 12.946ν24 = 15ν13 + 10ν25. Both sides equal 6.9757.Second equality: 12× 12.425ν12 + 24× 12.946ν24 = 13× 15ν13 + 25× 10ν25. Both sides equal 108.21.Third relation: lhs = 144×12.425ν12 +576×12.946ν24 = 1886.46 and rhs = 169×15ν13 +625×10ν25 = 1844.73.hence we have the conditions for immunisation against small interest rate changes.

6. (i) For medium and long-term investments. Unsecured. Bearer bonds. Usually one annual coupon. Issued ineurocurrency—currency owned by a non-resident of the country where the currency is legal tender.(ii)(a) Now 97 = ra20 ,0.05 + 100/1.0520. Hence r = (97− 100/1.0520)/a20 ,0.05 = 4.75927. (b) We have

dM (i) =1

97

[r

20∑k=1

k

1.05k+

100× 201.0520

]=

197

[r(Ia)20 ,0.05 +

100× 201.0520

]= 13.2146691547

So the duration is 13.2147 years.


7. (i) Let ν = 1/1.07.First condition: present values are equal. For the assets we have VA = 7.404ν2 + 31.834ν25 = 12.3323.For the liabilities we have VL = 10ν10 + 20ν15 = 12.3324.Second condition: the durations are the same.For the assets we have ∑ tkatk

(1 + i)tk= 2× 7.404ν2 + 25× 31.834ν25 = 159.569

For the liabilities we have ∑ tk`tk(1 + i)tk

= 10× 10ν10 + 15× 20ν15 = 159.569

Hence the durations (which are these figures multiplied by ν/P ) are equal.(ii) Let ν1 = 1/1.075. Then profit is 7.404ν2

1 + 31.834ν251 − 10ν10

1 − 20ν151 = 0.0158.

(iii) The assets are clearly more spread out than the liabilities; hence all three of Redington’s are satisfied and thecompany is immunised against small changes in the interest rate.

8. (i) Let ν = 1/1.07. The discounted mean term is

dM (i) =1P

∑ tkctk(1 + i)tk

=8× 87.5ν8 + 19× 157.5ν19

87.5ν8 + 157.5ν19 =700 + 2992.5ν11

87.5 + 157.5ν11 = 13.07061477

or 13.070615 years. The convexity is

c(i) =1P

∑ tk(tk + 1)ctk(1 + i)tk+2 =

8× 9× 87.5ν10 + 19× 20× 157.5ν21

87.5ν8 + 157.5ν19 =6300ν2 + 59850ν13

87.5 + 157.5ν11 = 186.8959854

or 186.895984.(ii) We need equal present values. Hence 87.5ν8 + 157.5ν19 = 66.85ν4 +Xνn. Hence Xνn = 43.4763142.We also need equal durations. Hence we need 8 × 87.5ν8 + 19 × 157.5ν19 = 4 × 66.85ν4 + nXνn. This secondequation is 700ν8 + 2992.5ν19 = 267.4ν4 + nXνn. Hence nXνn = 1030.85938.Hence n = 23.710827 and hence X = 43.4763142× 1.07n = 216.25511.Now we check the third condition: for the assets

∑t2katkν

tk = 16× 66.85ν4 + n2Xνn = 25258.52023.For the liabilities:

∑t2k`tkν

tk = 18980.82353. Hence immunisation occurs.

9. (i) Present value of liabilities is 4ν19 + 6ν21. The zero-coupon bond must have the same present value. Hence it musthave some value £X at time n years where 19 < n < 21. Hence its spread is less than the spread of the liabilitiesand so immunisation is not possible.(ii) Let ν = 1/1.07. Then present value of assets is VA = 3.43ν15 + 7.12ν25 = 2.5550 and present value of liabilitiesis VL = 4ν19 + 6ν21 = 2.5551. Equal to third decimal place.We also need equal durations. So we need: 15×3.43ν15 +25×7.12ν25 = 19×4ν19 +21×6ν21. Now lhs = 51.44420and rhs = 51.44528. Equal to second decimal place.Now we need to check convexity of assets is greater than convexity of liabilities. For the assets

∑t2katkν

tk =152 × 3.43ν15 + 252 × 7.12ν25 = 1099.6266. For the liabilities

∑t2k`tkν

tk = 192 × 4ν19 + 212 × 6ν21 = 1038.3217.This implies convexity of assets is greater than convexity of liabilities and hence we have immunisation.

10. (i) Let ν = 1/1.08.Present value of liabilities is VL = 3ν3 +5ν5 +9ν9 +11ν11 = 15.0044. Present value of assets is VA = a5 ,0.08 +Rνn =3.992710 +Rνn. Equating these gives Rνn = 11.01169.We also need equal volatilities. For the liabilities,

∑tk`tkν

tk = 9ν3+25ν5+81ν9+121ν11 = 116.5741. For the assets,∑tkatkν

tk = (Ia)5 ,0.08 + nRνn. Hence nRνn = 116.5741 − (Ia)5 ,0.08 = 116.5741 − (a5 ,0.08 − 5ν6)/(1 − ν) =105.20899.Hence n = 9.5543 and hence R = 11.01169× 1.08n = 22.9718.For immunisation we need the convexity of the assets to be greater than the convexity of the liabilities. For the assets∑t2katkν

tk =∑5k=1 k

2νk + n2Rνn = 40.275 + n2Rνn = 1045.474. For the liabilities∑t2k`tkν

tk = 27ν3 + 125ν5 +729ν9 + 1331ν11 = 1042.031. This implies convexity of assets is greater than convexity of liabilities and hence wehave immunisation.(ii)(a) Let ν1 = 1/1.03. For the liabilities we have VL = 3ν3

1 + 5ν51 + 9ν9

1 + 11ν111 = 21.9029. For the assets we have

VA = a5 ,0.03+Rνn = 4.579707+22.9718ν9.5543 = 21.8996. This is a deficit of 0.0033 or £3,300. (b) Immunisationonly protects against small changes in the interest rate. This is a large change!

11. (i)(a) Let ν = 1/1.06. Present value of liabilities is VL = a20 ,0.06 + 0.5ν20a20 ,0.06 = (1 + 0.5ν20)a20 ,0.06. Durationof liabilities is

dM (i) =1VL

[(Ia)20 ,0.06 + 0.5(21ν21 + 22ν22 + · · · + 40ν40)

]=

0.5(Ia)40 ,0.06 + 0.5(Ia)20 ,0.06

(1 + 0.5ν20)a20 ,0.06

Using (Ia)n = (an − nνn+1)/(1− ν) and tables gives

dM (i) =a40 ,0.06 − 40ν41 + a20 ,0.06 − 20ν21

(1− ν)(2 + ν20)a20 ,0.06= 11.302655


or 11.30265 years.(b) Present value of assets for £100 nominal is VA = 10a15 ,0.06 + 100ν15. Hence duration of assets is

dM (i) =10(Ia)15 ,0.06 + 15× 100ν15

10a15 ,0.06 + 100ν15 =(Ia)15 ,0.06 + 150ν15

a15 ,0.06 + 10ν15 = 9.3523637

or 9.35236 years.(ii)(a) For immunisation, the duration of the assets must equal the duration of the liabilities. Part (i) shows that theduration of assets invested in the government bonds will always be too short. (b) If interest rates decrease, thenthe values of VA and VL would both rise. The duration of the liabilities is greater than the duration of the assets andthe duration is a measure of the rate of change of the present value with respect to i. Hence the value of VL wouldrise by more than the rise in VA.

12. (a) The cash flow (in £1m) is as follows:Time 0 1 2 3 4 5 6 7 8 9 10Cash flow −P 1 1 1 1 1 1 1 1 1 1

Let ν = 1/1.07. Then

dM (i) =ν + 2ν2 + · · · + 10ν10

P=

(Ia)10 ,0.07

a10 ,0.07=

(a10 ,0.07 − 10ν11)(1− ν)a10 ,0.07

= 4.946

(b) There are 3 conditions to check:(I)∑nk=1 ltkν

tk = 7ν5 + 8ν8 = 9.646976 and∑nk=1 atkν

tk = a10 ,0.07 +Xνn = 7.023582 +Xνn.Hence we want Xνn = 2.623394.(II)∑nk=1 tkltkν

tk = 7× 5ν5 + 8× 8ν8 = 62.2031∑nk=1 tkatkν

tk = (ν + 2ν2 + · · · + 10ν10) + nXνn = (Ia)10 + nXνn = (a10 ,0.07 − 10ν11)/(1− ν) + nXνn.Hence n = 10.46886 years and so X = 5.32695 million.(III)

∑nk=1 t

2kltkν

tk = 7× 25ν5 + 8× 64ν8 = 422.7612∑nk=1 t

2katkν

tk = (ν + 22ν2 + · · · + 102ν10) + n2Xνn = 228.451 + n2Xνn = 515.9673As∑nk=1 t

2katkν

tk >∑nk=1 t

2kltkν

tk , we have Redington immunisation.

13. (Ia)n is the present value (time 0) of the following increasing annuity:

Time 0 1 2 · · · nCash Flow, c 0 1 2 · · · n

Hence:(Ia)n = ν + 2ν2 + · · · + nνn

(1− ν)(Ia)n = ν + ν2 + · · · + νn − nνn+1 = νan − nνn+1

= ν(an − nνn)

and so

(Ia)n =ν

1− ν[an − nνn

]=

an − nνn

i=

an − nνn+1

1− ν(ii) The cash flow is as follows:

Time 0 0.5 1 · · · 9 10Cash Flow, c −P 5 5 · · · 5 105

Now

P = 5(ν + ν2 + · · · + ν20) + 100ν20 where ν =1

1.03= 5a20 ,0.03 + 100ν20 = 129.755

Hence

dM (i) =1P

n∑k=1

tkctk(1 + i)tk

=1P

[5(0.5ν + ν2 + 1.5ν3 + · · · + 10ν20) + 10× 100ν20]

=1P

[2.5(Ia)20 + 1000ν20] =

12P

[5a20 − 20ν21

1− ν+ 2000ν20

]=

1815.7392P

= 6.997 years

(iii) Now suppose the coupon rate is £8 per annum instead of £10 per annum. The duration is an average of the timesof payments weighted by their sizes: decreasing the coupon to £8 means that the final payment has relatively moreweight and so the duration will be longer.


14. (i) Let ν = 1/1.06. Ignoring tax we would have P = 4a(2)20 ,0.06 + 110ν20. But we have CGT at the rate of 25% and

income tax of an annual amount of 1 at the end of each year. Hence the price is given byP = 4a(2)

20 ,0.06 − a20 ,0.06 + [110− 0.25(110− P )] ν20

and hence

P =4a(2)

20 ,0.06 − a20 ,0.06 + 82.5ν20

1− 0.25ν20 =(4× 1.014782− 1)× 11.469921 + 82.5ν20

1− 0.25ν20 = 65.95296

(ii) Now there is no CGT. Hence the price is given byP = 4a(2)

20 ,0.06 − a20 ,0.06 + 110ν20

= (4× 1.014782− 1)× 11.469921 + 110ν20 = 69.38648We now need to evaluate

∑tkctkν

tk for the following cash flow:Time 0 0.5 1 1.5 2 2.5 3 · · · 19.5 20Cash Flow—coupons 0 2 2 2 2 2 2 · · · 2 2Cash Flow—redemption value 0 0 0 0 0 0 0 · · · 0 110Cash Flow—tax 0 0 −1 0 −1 0 −1 · · · 0 −1

Let x = ν1/2. Then∑tkctkν

tk = 2× 0.5(x + 2x2 + · · · + 40x40) + 20× 110x40 − (Ia)20 ,0.06

= (x + 2x2 + · · · + 40x40) + 2200ν20 − (Ia)20 ,0.06

= x + 2x2 + · · · + 40x40 + 587.2700

=x(1− x40)(1− x)2 −

40x41

1− x+ 587.2700 = 976.09822

and hence

dM (i) =976.0982269.38648

= 14.067556

or 14.0676 years.

15. Use units of £1,000.(i) Let ν = 1/1.07. The present value of the liabilities is VL = 160a15 ,0.07 + 200ν10 = 1558.93610.(ii) Discounted mean term is

dM (i) =∑tkctkν

tk

VL=

160(Ia)15 ,0.07 + 10× 200ν10

160a15 ,0.07 + 200ν10 =16(Ia)15 ,0.07 + 200ν10

16a15 ,0.07 + 20ν10 = 6.96971

or 6.96971 years.(iii) Let A denote the nominal amount of security A purchased and B denote the nominal amount of security Bpurchased. Then the present value of the assets is VA = A(0.08a8 ,0.08 +ν8)+B(0.03a25 ,0.08 +ν25) and the discountedmean term is

A(0.08(Ia)8 ,0.08 + 8ν8) +B(0.03(Ia)25 ,0.08 + 25ν25)VA

So we have the equations1.05971302A + 0.53385667B = 1558.936106.63689237A + 7.97613131B = 10865.33253

Discriminant is ∆ = 6.63689237 × 0.53385667 − 1.05971302 × 7.97613131 = −4.90926094. Hence A =(10865.33253 × 0.53385667 − 1558.93610 × 7.97613131)/∆ = −6633.74879459/∆ = 1351.272396 and B =(6.63689237 × 1558.93610 − 1.05971302 × 10865.33253)/∆ = −1167.64324/∆ = 237.8450. So we need£1,351,272.40 of A and £237,845.00 of B.(iv) The assets are more spread out than the liabilities. Hence Redington immunisation should hold.

16. Let ν = 1/1.07 and x = ν5. Liabilities:Time 0 5 10 15 20 25Cash Flow 0 3,000 5,000 7,000 9,000 11,000

(i) NPV = 1000(3ν5 + 5ν10 + 7ν15 + 9ν20 + 11ν25) = 1000x[3 + 5x + 7x2 + 9x3 + 11x4] = 11,570.339.(ii) dM (i) = 5000x[3 + 10x + 21x2 + 36x3 + 55x4]/NPV = 171,353/NPV = 14.8097 years.(iii) The present value of £100 nominal of A is 5a26 ,0.07 + 100ν26 = 76.348443. Also the present value of £100nominal of B is 4a32 ,0.07 + 100ν32 = 62.06033405.Equating the present values gives one equation: 76.348443A + 62.06033405B = 11,570.339.For dM (i) for the assets, we proceed as follows:

5(Ia)26 ,0.07 + 100× 26ν26 = 5

[a26 ,0.07 − 26ν27

1− ν

]+ 2600ν26 = 1031.744022


Similarly:

4(Ia)32 ,0.07 + 100× 32ν32 = 4

[a32 ,0.07 − 32ν33

1− ν

]+ 3200ν32 = 930.6057861

Hence the second equation is 1031.744022A + 930.6057861B = 171,353. Solving these 2 equations gives: A =18.9746 and B = 163.09. Hence we need £1,897.46 nominal of A and £16,309 nominal of B.

17. (i) Let ν = 1/1.06, α = 1.1ν and β = 1.03ν. Then for Cyber plc we have

P = 6ν6 + 6× 1.1ν7 + 6× 1.12ν8 + · · · + 6× 1.16ν12 +∞∑k=1

(6× 1.16 × 1.03kν12+k)

= 6ν6(1 + α + α2 + · · · + α6) + 6ν6α6∞∑k=1

1.03kνk = 6ν6[

1− α7

1− α+ α6 β

1− β

]= 214.544

and for Boring plc we have

P = 4ν(1 + 1.005ν + 1.0052ν2 + · · ·) =4ν

1− 1.005ν= 72.727

(ii) Now let ν = 1/1.07. For Cyber plc we have:

P = 6ν6[

1− α7

1− α+ α6 β

1− β

]= 151.982 Drop is 29.16%.

and for Boring plc we have

P =4ν

1− 1.005ν= 61.538 Drop is 15.38%.

(iii) The cash flows for Cyber plc are later. Hence its duration is larger and hence its dependence on the interest rateis larger.

18. (i)(a) The volatility or modified duration is given by

d(i) = − 1P (i)

dP

di=

n∑k=1

tkCtk(1 + i)tk+1

/n∑k=1

Ctk(1 + i)tk

(b) Convexity is

c(i) =1P

d2P

di2=

n∑k=1

tk(tk + 1)Ctk(1 + i)tk+2

/n∑k=1

Ctk(1 + i)tk

(ii)(a) Using 100ian,i + 100νn = 100 gives volatility is (8ν(Ia)10 ,0.08 + 1000ν11)/100 = a10 ,0.08 = 6.710081.(b) Volatility: 100,000× 7.247× ν8.247/(100,000ν7.247) = 7.247ν = 6.71018.Convexity: 100,000× 7.247× 8.247ν9.247/(100,000ν7.247) = 7.247× 8.247ν2 = 51.240.(c) Present value of liability of £100,000 to be paid in 7.247 years is 100,000ν7.247 = 57,250.337. This is the amountinvested in the loan stock.Hence NPV of assets = NPVof liabilities = 57,250.337.Volatility of assets = volatility of liabilities, by (ii) (a) and (b)Convexity of assets = 60.53 and convexity of liabilities = 51.24 by (ii)(b).Hence Redington immunisation.(iii) Liabilities: 100,000 to be paid in 7.247 years.Assets: 57,250.337 of loan stock with 8% couponNPV of liabilities = 100,000/1.0857.247 = 55,365.685.NPV of assets = 57,250.337(0.08a10 ,0.085 + ν10) = 55,372.14. Hence profit (in 1/3/1997 values) is 6.45.

19. Let ν = 1/1.07. The cash flows for the annuity certain are:Time 0 · · · 9.5 10 10.5 11 · · · 34 34.5Cash flow 0 · · · 0 2500 2500 2500 · · · 2500 2500

Value of this annuity at time 9.5 is a(2)25 ,0.07.

Hence, net present value of liabilities is

PL = 20ν15 + 5ν9.5a(2)25 ,0.07 = 20ν15 + 5ν9.5 i

i(2) a25 ,0.07

= 20ν15 + 5ν9.5 × 1.017204× 11.653583 = 38.41568If Vl denotes the volatility of liabilities, then

VL × PL = 20× 15ν16 + 2.5(10ν11 + 10.5ν11.5 + · · · + 34.5ν35.5)

= 300ν16 + 1.25ν10.5(20ν0.5 + 21ν1 + · · · + 69ν25.0) = 300ν16 + 1.25ν10.550∑k=1

(k + 19)νk/2

Let x = ν1/2. Then


VL × PL = 300x32 + 1.25x2150∑k=1

(k + 19)xk

= 300x32 + 1.25x21(

19x1− x50

1− x+x(1− x50)(1− x)2 −

50x51

1− x

)= 300ν16 + 1.25x21

(19x

1− ν25

1− x+x(1− ν25)(1− x)2 −

50x51

1− x

)= 101.62038 + 1.25x21(450.45473 + 712.73726− 267.74145) = 651.69543

Assets: suppose invest in yY nominal of bond Y and it matures at time tY .Present value of assets:

25ν10 + yY νtY = PL = 38.41568(i) Hence yY νtY = 25.70695.(ii) Volatility of assets:

25× 10ν11 + tY yY νtY +1 = VL × PL = 651.69543and hence tY yY νtY +1 = 532.922.Hence tY = 22.182 years and yY = 115.302.(iii) For Redington immunisation, have 3 conditions:• value of assets = value of liabilities• volatility of assets = volatility of liabilities• convexity of assets > convexity of liabilitiesFirst two conditions are satisfied. Remains to check last condition, where convexity is given by:

c(i) =n∑k=1


/P

20. (i)(a) and (b) in notes.(ii) Present value of liabilities is:

PL = 60a9 ,0.07 +750

1.0710 = 772.176

Volatility of liabilities is:

VL =60(Ia)9 ,0.07 + 10× 750ν10

PLLet yA and yB denote nominal amounts in bonds A and B respectively. Then net present value of assets is

yAν5 + yBν20 = PL

The volatility of the assets is5yAν5 + 20yBν20

PLThis leads to yB = 115.406 or £115,406 and yA = 656.768 or £656,768.(iii) For Redington immunisation, have 3 conditions:• value of assets = value of liabilities• volatility of assets = volatility of liabilities• convexity of assets > convexity of liabilitiesFirst two conditions are satisfied. Remains to check last condition, where convexity is given by:

c(i) =n∑k=1


/P

21. (i)(a) Present value of liabilities is VL = 100a5 ,0.05 = 432.9477.(b) Let ν = 1/1.05. Macaulay duration is dM (i) = 100(Ia)5/VL = 100(a5 − 5ν6)/(1− ν)VL = 2.90 years.(ii) Suppose purchase nominal yA of 1-year bond and yB of 5-year bond.Then present value is yAν + yBν5 = VL. Macaulay duration is (yAν + 5yBν5)/VL = dM (i). Hence (yAν + 5yBν5) =VLdM (i) = 100(a5 − 5ν6)/(1− ν) = 1256.64. Hence 4yBν5 and so yB = 262.8158. Hence yA = 238.3759.(iii) (a) Convexity of assets: (2yAν3 + 30νBν7)/(yAν + yBν5) = 13.89356 or 13.89 years. (b) Assets more spreadout than liabilities. Hence Redington’s immunisation should have been achieved.Check of convexity of liabilities: 100(2ν + 6ν2 + 12ν3 + 20ν4 + 30ν5)/432.9477 = 12.08. Hence convexity of assetsis greater than convexity of liabilities.

22. For proof of first formula, see exercise 10 in section 3.3.Liabilities (in 1,000s):

Time 0 1 2 3 · · · 19 20Cash flow 0 100 105 110 · · · 190 195


(a) Present value: NPV = 95a20 ,0.07 + 5(Ia)20 ,0.07 = 95a20 ,0.07 + 5(a20 ,0.07 − 20ν21)/(1− ν) = 1446.947.Duration: dM (i) =

∑20k=1(95 + 5k)kνk/NPV = (95(Ia)20 ,0.07 + 5

∑20k=1 k

2νk)/NPV = 13643.889/NPV = 9.429 or9.43 years.(b) Suppose xA and xB denote nominal (or face values) of amounts on (A) and (B) respectively. Hence present valueis xAν25 + 0.08xBa12 ,0.07 + xBν12 = 1446.947. Duration is (25xAν25 + 0.08xB(Ia)12 ,0.07 + 12xBν12)/1446.947 =9.429. We have two simultaneous equations for the two unknowns: xA and xB . Solving (take 25 × first equationminus second equation) gives xB = 1249.27 and hence xA = 534.27.

23. Cash flow:Time 0 1/2 1 3/2 4/2 · · · 39/2 40/2Cash flow 0 3.75 3.75 3.75 3.75 · · · 3.75 113.75

(i)(a) P = 7.5a(2)20 ,0.1 + 110ν20 = 7.5(i/i(2))a20 ,0.1 + 110ν20 = 7.5× 1.024404× 8.513564 + 110/1.120 = 81.76077.

(b) Let dM (i) denote the volatility. Hence

dM (i)× P =∑ tkCtk

(1 + i)tk+1 =40∑k=1

k

23.75νk/2+1 + 20× 110ν21 =

3.75ν2

40∑k=1

kνk/2 + 2,200ν21

=3.75ν

2

2a(2)20 − 40ν41/2

1− ν1/2 + 2,200ν21

and hence dM (i) = 8.9104.(ii) Let x denote amount of each liability. Let t1 denote time first liability is due. Then present value of liabilities isx(νt1 + ν10) = P = 81.76077.Let dLM denote volatility of the liabilities. Then dLM × P = x(t1νt1+1 + 10ν11).Hence (t1νt1+1 + 10ν11)/(νt1 + ν10) = 8.9104 and substituting t1 = 9.61 shows that this is the required value.(b) Convexity:

c(i) =n∑k=1


/P =

t1(t1 + 1)xνt1+2 + 10× 11xν12

xνt1 + xν10 = 87.526

Payments which are more spread out will have greater convexity. In this case, liabilities are very close (9.61 and 10years) and assets more spread out. Hence there should be immunisation.(iii) Present value one year later (at time t = 1) is 7.5a(2)

19 ,0.1 + 110ν19 = 7.5 × 1.024404 × 8.36492 + 110/1.119 =82.254.Present value (times t = 1) of interest payments for next four years is 7.5a(2)

4 ,0.1 = 24.354.Hence present value (time t = 1) of asset minus interest payments is 57.9. Hence forward price at time 5 (four yearslater) is 57.9× 1.14 = 84.77.

24. (i) Let ν = 1/1.03, α = 1.05 and β = αν = 1.05/1.03. Then

NPV = 100[ν + αν2 + α2ν2 + · · · + α59ν60] =

100ν(1− β60)1− β

= 10852.379704

(ii) Let x denote the nominal amount of the bond. Then for the assets

NPV = 0.04xa20 ,0.03 +x

1.0320 = x [0.04× 14.877475 + 0.553676]

Hence x = 9446.91494 or £9,446.91494m.(iii) Duration of liabilities:

dM (i) =1

NPV

∑ctktkν

tk =100NPV

[ν + 2αν2 + 3α2ν3 + · · · + 59α58ν59 + 60α59ν60]

=100νNPV

[1 + 2β + 3β2 + · · · + 59β58 + 60β59] =

100νNPV

[1− β60

(1− β)2 −60β60

1− β

]= 36.143707

or 36.1437 years.(iv) Duration of the assets:

1NPV

[0.04x(1 + 2ν + 3ν2 + · · · + 20ν20) + 20xν20] =

x

NPV

[0.04ν

(1− ν20

(1− ν)2 −20ν20

1− ν

)+ 20ν20

]= 14.572532

or 14.57 years.(v) For the liabilities, d(i) = νdM (i) = 35.090977 and for the assets d(i) = 14.148089. Hence if there is a change of1.5% in the interest rate, then the liabilities change by 35× 1.5 = 52.5% approximately whilst the assets change by14.15 × 1.5 = 21.2% approximately. Thus the liabilities would increase by approximately 52.5 − 21.2 = 31.3% oftheir original value more than the change in the value of the assets.


25. (i)(a) Now P = 500a20 ,0.08. Hence dM (i) = (Ia)20/

a20 = (1.08/0.08) × (9.818147 − 20/1.0821)/9.818147 =8.03694757 or 8.037 years. (i)(b) Let α = 1.08 and ν = 1/1.08. Then P = 500ν(1 + αν + · · · + α19α19) =500ν × 20 and dM (i) = 500ν(1 + 2αν + · · · + 20α19α19)/P = (1 + 2 + 3 + · · · + 20)/20 = 10.5 or 10.5 years.(ii) The duration is a measure of the mean term of the cash flows weighted by their present values. For (b), the latercash flows are higher than the corresponding cash flows in (a); hence the duration will be higher.

26. Let ν = 1/1.07 and α = 1.05.First option.

Outgoings = 0.25 + 0.1ν + 0.2ν2 + · · · + 0.9ν9 + ν10 = 0.25 +1

10(ν + 2ν2 + · · · + 9ν9 + 10ν10)

= 0.25 +1

10

[ν(1− ν10)(1− ν)2 −

10ν11

1− ν

]Incomings = 0.5(ν8 + ν9 + · · ·+ ν27) + 5ν27 = 0.5ν8(1− ν20)/(1− ν) + 5ν27. Then NPV = Incomings−Outgoings =0.379448 or £379,448.Second Option. NPV = 0.21ν[1+αν+α2ν2+· · ·+α9ν9]+5.64ν10−4.2 = 0.21ν(1−α10ν10)/(1−αν)+5.64ν10−4.2 =0.472592 or £472,592. Hence the second option has the larger NPV.(ii) Recall the discounted mean term or Macaulay duration is mean term of the cash flows weighted by their presentvalues: dM (i) =

∑tkctkν

tk/P . Clearly the discounted mean term of the second option is less than 10 years. Thediscounted mean term of the first option, d1 is given by d1P = −0.1(ν + 22ν2 + 32ν3 + · · ·+ 102ν10) + 0.5(8ν8 + 9ν9 +· · · + 27ν27) + 5 × 27ν27. The sum over the first 10 years (terms up to ν10) is negative. Hence the discounted meanterm must be larger than 10 years. (In fact, d1P = 49.404532.)

27. Let ν = 1/1.08. Use units of £1,000.Present value of liabilities is VL = 400ν10. Hence, amount held in cash is 40ν10.Let x denote nominal amount in the zero-coupon bond and let y denote the nominal amount in the fixed-intereststock. Hence the present value of the assets is VA = 40ν10 + xν12 + 0.08ya16 ,0.08 + 1.1yν16.Second condition is

∑k tkatkν

tk =∑k tk`tkν

tk . The right hand side is 4,000ν10 and the left hand side is 12xν12 +0.08y(Ia)16 ,0.08 + 1.1× 16yν16. This gives two equations for the two unknowns, x and y:

xν12 + 0.08ya16 ,0.08 + 1.1yν16 = 360ν10

12xν12 + 0.08y(Ia)16 ,0.08 + 17.6yν16 = 4,000ν10

Multiplying the first equation by 12 and subtracting the second equation givesy[0.96a16 ,0.08 − 4.4ν16 − 0.08(Ia)16 ,0.08

]= 320ν10

and hence y = 63.785239 or £63,785. Substituting in the first equation gives x = 254.593513 or £254,593.(ii) Amount invested in the zero-coupon bond is xν12 = 101.102587 or £101,103. Hence amount in the fixed intereststock must be 400ν10 − 40ν10 − xν12 = 0.08ya16 ,0.08 + 1.1yν16 = 65.647069 or £65,647.(iii) The assets are more spread out than the liabilities (which are at a single point). Hence we do have immunisation.

28. Let ν = 1/1.08 and work in units of £1,000. (i) We have VL = 300ν15 + 200a20 ,0.08 = 2058.20191149 or£2,058,201.91. (ii) We have

dM (i) =∑tk`tkν

tk

VL=

15× 300ν15 + 200(ν + 2ν2 + · · · + 20ν20)VL

=4,500ν15 + 200(Ia)20 ,0.08

VL= 8.356894

or 8.36 years. (iii) Suppose purchase yA of security A and yB of security B. Then VA = yA(0.09a12 ,0.08 + ν12) +yB(0.04a30 ,0.08 + ν30) = 1.07536078yA + 0.54968865yB . We also need

∑tkatkν

tk = 4,500ν15 + 200(Ia)20 ,0.08 =17200.17420214 where∑

tkatkνtk = yA

[0.09(Ia)12 ,0.08 + 12ν12] + yB

[0.04(Ia)30 ,0.08 + 30ν30] = 8.56066413yA + 7.56986281yB .

So we have two equations in two unknowns:1.07536078yA + 0.54968865yB = 2058.201911498.56066413yA + 7.56986281yB = 17200.17420214

and hence

yA =7.56986281× 2058.20191149− 0.54968865× 17200.17420214

1.07536078× 7.56986281− 8.56066413× 0.54968865= 1783.469843361

Similarly, yB = 255.2870422689. So we need £1,783,470 in security A and £255,287 in security B. (iv) We needthe convexity of the assets to be greater than the convexity of the liabilities. Assuming the first two conditions holdas in part (iii), this reduces to: ∑

t2katkνtk >

∑t2k`tkν

tk

29. (i) See notes. (ii) Recall duration is dM (i) =∑tkctkν

tk/P . We have ν = 1/1.05.One year bond: P = 104ν and dM (i) = 104ν/P = 1 or 1 year.Three year bond: P = 4ν + 4ν2 + 104ν3 and dM (i) = (4ν + 8ν2 + 312ν3)/P = 2.88437972 or 2.88 years.Five year bond: P = 4a5 ,0.05 + 100ν5 = 95.67052465 and dM (i) = (4(Ia)5 ,0.05 + 500ν5)/P . Now (Ia)5 ,0.05 =(a5 ,0.05 − 5ν6)/(1− nu). Hence dM (i) = 4.62032257 or 4.62 years.


(iii) The duration, dM (i), is the average of the times tk weighted by their present values, ctkνtk/P . If the coupon

rate increased to 8%, then the coupons would form a larger part of the return from the bond. Hence the average timeof the cashflows would be less.(iv) Let i denote the rate of return and ν = 1/(1 + i).Option 1: we have 95 = 4a4 ,i + 79ν5. Because 95 = 4 + 4 + 4 + 4 + 79, the rate of return is clearly 0.Option 2: the new cashflow is as follows

time 0 1 2 3 4 5 6 7 8 9 10 11 12cashflow -95 4 4 4 4 1 1 1 1 1 1 1 101

Hence 95 = 4a4 ,i + ν4a8 ,i + 100ν12. A capital gain of 6 is obtained after 12 years; this is an average of 0.5 per year.So that suggests the rate of return is between 1.5% and 4.5% and closer to 1.5% than to 4.5%.Trying i = 0.03 gives 95− 4a4 ,i − ν4a8 ,i − 100ν12 = 3.77426736.Trying i = 0.025 gives 95 − 4a4 ,i − ν4a8 ,i − 100ν12 = −0.89927474. Hence required i = 0.025 + 0.005 ×0.89927474/(3.77426736 + 0.89927474) = 0.02596209 or 2.6% per annum.(v) Harder to predict inflation risk for Option 2. Tax will be different for the two options. Risk of further defaultharder to predict for option with the longer term. Cash obtained from Option 1 could be invested elsewhere toimprove return from Option 1 over the period of 12 years.

30. (i) The payback period is the time when the net cash flow is positive—it completely ignores the effect of interest.The DPP is the time, t, when the NPV of payments up to time t is positive. However, it does not look at the size ofthe overall profit of the project.(ii)(a) Measure time in years. Then incomings over [0, t] are 0.64t. Hence we want t0 with 0.64t0 = 3; hencet0 = 4.6875 years. (b) Let ν = 1/1.04. Then NPV =

∫ρ(t)ν(t) dt = 0.64

∫ t0

0 νt dt = 0.64(νt0 − 1)/ ln ν. Settingthis equal to 3 gives νt0 = 3 ln ν/0.64 + 1; taking logs gives t0 = 5.17975169 or 5.18 years.(iii) Let α = 1.1. For project A we have NPV = fA(i) = −3 + 0.5ν1/2 + 0.5αν3/2 + · · · + 0.5α5ν11/2 = −3 +0.5ν1/2(1− α6ν6)/(1− αν). For project A we have NPV = fB(i) = −3 + 0.64

∫ 60 ν

t dt. When i = 0 we have ν = 1and fA(0)− fB(0) = −3 + 0.5(1− α6)/(1− α) + 3− 3.84 > 0. When i = 0.04 we have fA(0)− fB(0) < 0. Hencethere is at least one root in (0, 0.04).(iv) Duration of incomings of projectA is dAM (i) = ν1/2(1+3αν+5α2ν2 +· · ·+11α5ν5)/(4P ) where P = 0.5ν1/2(1−α6ν6)/(1−αν). Hence dAM (i) = 0.5(1+3αν+5α2ν2 + · · ·+11α5ν5)(1−αν)/(1−α6ν6) = 3.16327775 or 3.16 years.Duration of incomings of project B is dBM (i) =

∫ 60 0.64tνt dt/

∫ 60 0.64νt dt =

∫ 60 tν

t dt/∫ 6

0 νt dt. The denominator

is (ν6 − 1)/ ln ν and the numerator is 6ν6/ ln ν − (ν6 − 1)/(ln ν)2. Hence dBM (i) = 2.882446 or 2.88 years.(v) The duration is a measure of the sensitivity of the price to interest rate changes. The duration of the incomingsof project A is greater than the duration of the incomings of project B. Hence if the interest rate increases, thenet present value of the incomings of project A will decrease more than the net present value of the incomings ofproject B.

31. Let ν = 1/1.08. Now VL = 6ν8 + 11ν15 and VA = Xν5 + Y ν20. First two conditions are 6ν8 + 11ν15 = Xν5 + Y ν20

and 48ν8 + 165ν15 = 5Xν5 + 20Y ν20. Solving gives 15Xν5 = 72ν8 + 55ν15 and 15Y ν20 = 18ν8 + 110ν15. HenceX = 5.50877088 and Y = 13.7968767. (ii) For the third condition, we need to check

∑t2katkν

tk >∑t2kltkν

tk .Now lhs = 25Xν5 + 400Y ν20 = 1277.76749 and rhs = 6 × 48ν8 + 11 × 225ν15 = 935.820659. Hence the /thirdcondition is satisfied.

32. Let ν = 1/1.04. Use units of £1 million. The NPV of the liabilities is 10a40 ,0.04 = 197.92774 or £197,927,400.Let x denote the amount invested in the first security; then this security has NPV equal to 0.05xa10 ,0.04 + xν10 =1.081108969x. Hence x = 98.96387/1.081108969 = 91.5392183746 or £91,539,218.37.Let y denote the amount invested in the first security; then this security has NPV equal to 0.1ya5 ,0.04 + yν5 =1.267109307y. Hence y = 98.96387/1.267109307 = 78.102078055 or £78,102,078.06.(iii) The duration of the liabilities is 10(ν + 2ν2 + 3ν3 + · · · + 39ν39 + 40ν40)/197.92774 = 10(Ia)40 ,0.04/197.92774where (Ia)40 ,0.04 = (a40 ,0.04 − 40ν41)/(1 − ν) = 306.32307934. Hence the duration of the liabilities is 15.476511or 15.48 years.(iv) Duration of the assets is

[0.05x(Ia)10 ,0.04 + 10xν10 + 0.1y(Ia)5 ,0.04 + 5yν5

]/197.92774 where (Ia)10 ,0.04 =

(8.110896 − 10ν11)/(1 − nu) = 41.9922537936 and (Ia)5 ,0.04 = (4.451822 − 5ν6)/(1 − nu) = 13.0064836551.This leads to 6.23034183924 or 6.23 years.(v) The duration of the liabilities is much greater than the duration of the assets. Hence if the interest rate falls thenthe increase in the NPV of the liabilities will be more than the increase in the NPV of the assets. Hence the fund willmake a loss.

33. (i) The price is 9a10 ,0.07 + 100/1.0710 = 114.047167 or 114.047. (ii) Now dM (i) =∑nk=1 tkctkν

tk/114.047 =[9(Ia)10 ,0.07 + 1000ν10

]/114.047 = 7.19880 or 7.199 years. (iii) Duration is a measure of the position of the

mean of the cash flows. So if 3%, the mean will be later because of the payment of 100 at the end. (iv)(a) Effectiveduration d(i) = dM (i)/(1 + i) = 7.1988/1.07 = 6.728 years. (b) The effective duration is a measure of the rate ofchange of the net present value with i. If the interest rate changes from 0.07 to 0.07 + ε then the new price will beapproximately (1− εd(i))P (i) = (1− 6.728ε)114.047.


34. Recall the duration is dM (i) =∑nk=1 tkctkν

tk . For this question, ν = 1/1.04 = 25/26. (i) Now P = 500a12 ,0.04 anddM (i) = 500(Ia)12 ,0.04/500a12 ,0.04 = (9.385074 − 12ν13) × 1.04/0.04 × 1/9.385074 = 56.632809/9.385074 =6.0343487 or 6.03435 years. (ii) dM (i) =

[5(Ia)8 ,0.04 + 800ν8

]/[5a8 ,0.04 + 100ν8

]. Now denominator = 5 ×

6.732745 + 100ν8 = 106.7327455 and numerator = 5 × (6.732745 − 8ν9) × 1.04/0.04 + 800ν8 = 729.1188090.Hence dM (i) = 6.8312569 or 6.83126 years. (iii) The duration of the assets is greater than the duration of theliabilities. Now the duration is a measure of the rate of change of the NPV. Hence, if the interest rate increases by asmall amount, the NPV of the assets will decrease by a larger amount than the decrease in the NPV of the liabilities;hence the insurance company makes a loss.


1. Y = ln(1 + it) ∼ N (µ = 0.06, σ = 0.01). Hence Pr[it ≤ 0.05] = Pr[Y ≤ ln(1.05)] = Pr[(Y − 0.06)/0.01 ≤(ln(1.05)− 0.06)/0.01] = Φ(−1.12) = 1− Φ(1.12) = 1− 0.8686 = 0.1314

2. Accumulated value is A(12) = 10(1 + i1)(1 + i2) · · · (1 + i12) where i1, i2, . . . , i12 are i.i.d. with mean µ = 0.06 andstandard deviation σ = 0.01.Hence E[A(12)] = 10 (E[1 + i1])12 = 10(1 + µ)12 = 10× 1.0612 = 20.122Now E[A(12)2] = E[100(1 + i1)2(1 + i2)2 · · · (1 + i12)2] = 100

(E[(1 + i1)2]

)12= 100(σ2 + 1.062)12.

Hence var[A(12)] = E[A(12)2] − E[A(12)]2 = 100[(σ2 + 1.062)12 − 1.0624

]and so the standard deviation is√

var[A(12)] = 0.65775.

3. Let i1, i2 and i3 denote the annual effective rate of interest in years 2004, 2005 and 2006 respectively. Then theaccumulated value is for £100 nominal is

A = 109 + 4(1 + i3) + 4(1 + i2)(1 + i3) + 4(1 + i1)(1 + i2)(1 + i3)

Using independence givesE[A] = 109 + 4(1 + E[i3]) + 4(1 + E[i2])(1 + E[i3]) + 4(1 + E[i1])(1 + E[i2])(1 + E[i3])

= 109 + 4× 1.045 + 4× 1.045× 1.06 + 4× 1.045× 1.06× 1.055= 109 + 4.18[1 + 1.06 + 1.06× 1.055] = 122.285294

or £122,285.29.

4. Let S10 = (1 + i1) · · · (1 + i10) where i1, . . . , i10 are i.i.d. lognormal(µ = 0.075, σ2 = 0.00064).Hence lnS10 ∼ N (0.75, 0.0064).We want Pr(2000S10 > 4500) = Pr(S10 > 2.25) = Pr(lnS10 > ln 2.25) = Pr((lnS10 − 0.75)/0.08 > (ln 2.25 −0.75)/0.08) = 1− Φ(0.7616) = 1− (0.7764 + 0.16× 0.003) = 1− 0.7769 = 0.2231

5. Now 1 + i ∼ lognormal(µ, σ2) where E[1 + i] = 1.001 and var[1 + i] = 4× 10−6. We need to find µ and σ2. UsingE[Y ] = eµ+ 1

2σ2

= 1.001 and var[Y ] = e2µ+σ2(eσ

2 − 1) = 4 × 10−6 gives (eσ2 − 1) = 4 × 10−6/1.0012 and hence

eσ2

= 1 + 4× 10−6/1.0012 and σ2 = 3.992× 10−6. Hence µ = ln(1.001/eσ2/2) = 0.0009975. Or, quoting the results

of example 5.3b gives σ2 = ln(1 + 4 × 10−6/1.0012) = 3.9920 × 10−6 and µ = ln(1.0012/√

4× 10−6 + 1.0012) =0.0009975.Now Z = ln(1+ i) ∼ N (µ, σ2). We want j with P[i ≤ j] = 0.05. Hence 0.05 = P[1+ i ≤ 1+j] = P[ln(1+ i) ≤ ln(1+j)] = P[Z ≤ ln(1 + j)] = P[(Z−µ)/σ ≤ (ln(1 + j)−µ)/σ] = Φ((ln(1 + j)−µ)/σ). Hence ln(1 + j) = µ−1.644 854σgiving j = −0.002288.

6. Now S3 = (1 + i1)(1 + i2)(1 + i3) where i1, i2 and i3 are i.i.d. with expectation µ = 0.08 × 0.625 + 0.04 × 0.25 +0.02× 0.125 = 0.0625 and variance σ2 = (0.082 × 0.625) + (0.042 × 0.25) + (0.022 × 0.125)− µ2 = 0.00054375Now E[S3] = (1 + µ)3 = 1.06253 = 1.1995.Now E[(1 + i1)2] = var[1 + i1] + (1 + µ)2 = σ2 + (1 + µ)2 and hence E[S2

3 ] = (σ2 + (1 + µ)2)3. Hence var[S3] =(σ2 + (1 + µ)2)3 − E[S3]2 = (0.00054375 + 1.06252)3 − 1.06256 = 0.0020798 and so σ = 0.0456.

7. Let R = ln(1 + i) where E[i] = 0.0015 and var[i] = 0.0032 = 0.000009. Then R ∼ N (µ, σ2). Hence 1.0015 =E[1 + i] = E[eR] = eµ+ 1

2σ2

and 0.0032 = var[i] = var[eR] = e2µ+σ2(eσ

2 − 1). Hence eσ2 − 1 = 0.0032/1.0152 and so

σ2 = 8.9730× 10−6. Hence µ = ln 1.0015− 12σ

2 = 0.00149439.We want j with P[i > j] = 0.9. Hence 0.9 = P[i > j] = P[ln(1+ i) > ln(1+j)] = P[R > ln(1+j)] = P[(R−µ)/σ >(ln(1 + j)− µ)/σ)] = 1− Φ((ln(1 + j)− µ)/σ). Hence ln(1 + j) = µ + σΦ−1(0.1) = −0.0023445 or -0.23445%.

8. (i) Now Sn = (1 + i1)(1 + i2) · · · (1 + in). Using independence gives E[Sn] = Πnk=1 (1 + E[ik]) = (1 + j)n.

Now E[(1 + ik)2] = 1 + 2j + s2 + j2. Using independence again gives E[S2n] = (1 + 2j + s2 + j2)n and hence

var[Sn] = (1 + 2j + s2 + j2)n − (1 + j)2n.(ii)(a) E[S8] = (1 + j)8 = 1.068 = 1.593848. (b) var[S8] = (1.12 + 0.082 + 0.062)8 − 1.0616 = 0.343646.


9. Work in units of £1 million.(a) Let Rk denote the annual effective rate of return in year k Then E[Rk] = 1.06 and var[Rk] = 0.082 = 0.0064.Then S10 = R1R2 · · ·R10 and E[S10] = 1.0610 = 1.790848.(b) Now Rk is lognormal. Hence lnRk ∼ N (µ, σ2) where 1.06 = eµ+ 1

2σ2

and 0.0064 = e2µ+σ2(eσ

2 − 1). Usingthe method of example 5.3b gives σ2 = ln(1 + 0.0064/1.062) = 0.00567982 and µ = ln(1.062/

√0.0064 + 1.062) =

0.0554290.Now S10 ∼ lognormal(10µ, 10σ2). Hence P[S10 < 0.9E[S10]) = P[S10 < 1.611763] = P[ln(S10) < ln(1.611763)] =Φ((ln(1.61173)− 10µ)/

√10σ2) = Φ(−0.32301) = 0.3733.

10. Suppose 1 + it ∼ lognormal(µ, σ2) and hence Z = ln(1 + it) ∼ N (µ, σ2). Using E[etZ] = etµ+ 12 t

2σ2gives E[1 + it] =

eµ+ 12σ

2= 1.05 and var[1 + it] = e2µ+σ2

(eσ2 − 1) = 0.112. Hence eσ

2= 0.112/1.052 + 1 and so σ2 = 0.010915 and

µ = 0.0433325.(ii) Pr[0.04 < it < 0.07] = Pr[ln(1.04) < ln(1 + it) < ln(1.07)] = Φ((ln(1.07) − µ)/σ) − Φ((ln(1.04) − µ)/σ) =Φ(0.233)−Φ(−0.039) = Φ(0.233)− (1−Φ(0.039)) = Φ(0.233) + Φ(0.039)− 1 = 0.5910 + 0.3× 0.0038 + 0.5120 +0.9× 0.004− 1 = 0.1077

11. Let Z = 1+it. Then Y = lnZ ∼ N (µ, σ2). Then E[Z] = eµ+ 12σ

2and var[Z] = e2µ+2σ2

(eσ2−1). Hence eµ+ 1

2σ2

= 1.07and e2µ+σ2

(eσ2 − 1) = 0.04. Hence σ2 = ln(1 + 0.04/1.072) = 0.03434 and µ = ln(1.072/

√1.072 + 0.04) = 0.05049.

(ii) S15 = (1 + i1)(1 + i2) · · · (1 + i15) and hence lnS15 =∑15j=1 ln(1 + ij) ∼ N (15µ, 15σ2). Hence S15 ∼

lognormal(15µ, 15σ2).Pr[S15 > 2.5] = Pr[lnS15 > ln 2.5] = Pr[(lnS15 − 15µ)/

√15σ2 > (ln 2.5 − 15µ)/

√15σ2] = 1 − Φ((ln 2.5 −

15µ)/√

15σ2)) = 1− Φ(0.221457) = 1− 0.5877 = 0.4123.(Note that interpolation has been used: (Φ(0.221457)−Φ(0.22))/(Φ(0.23)−Φ(0.22)) = (0.221457− 0.22)/(0.23−0.22) = 0.1457 and hence Φ(0.221457) = Φ(0.22) + 0.1457(Φ(0.23)− Φ(0.22)) = 0.5877.)

12. We are given that ln(1 + it) ∼ N (µ = 0.06748, σ2 = 0.0003493) and E[1 + it] = 1.07 and var[1 + it] = 0.0004.(i) (a) 950,000× 1.05 = 997,500. Hence probability = 1.(b) 15% of assets in guaranteed return gives 950,000×0.15×1.05 = 149,625. Let S denote accumulated value from85% in the risky investment. Hence S = 950,000 × 0.85(1 + it) = 807,500(1 + it). We want Pr[S < 1,000,000 −149,625] = Pr[S < 850,375] = Pr[1 + it < 1.053096] = Pr[ln(1 + it) < ln 1.053096] = Pr[(ln(1 + it) − µ)/σ <(ln 1.053096− µ)/σ] = Φ(−0.8425) = 1− Φ(0.8425) = 1− (0.7995 + 0.0007) = 0.20(ii) (a) Variance = 0. (b) Return = 149,625 +S = 149,625 + 807,500(1 + it) which has variance 807,5002var(it) =807,5002 × 0.022 = 260,822,500. The units of the variance are £2.

13. Use units of £1,000.(i) A(5) = 425 × 1.035 + 425S5 where S5 = (1 + i1)(1 + i2) · · · (1 + i5). Using independence gives E[A(5)] =425×1.035 +425E[S5] = 425×1.035 +425×1.0355 = 997.4581615 or £997,458.16. Also var[A(5)] = 4252var[S5].Using independence again gives E[S2

5 ] = Π5k=1E(1 + ik)2 = (1 + 2 × 0.035 + 0.032 + 0.0352)5 = 1.4165344. Hence

var[S5] = 1.4165344− 1.03510 = 0.0059356. Hence the standard deviation of A(5) is 425√

var[S5] = 32.7432092or £32,743.21.(ii) The new expectation is 850 × 1.0355 = 1009.53336 or £1,009,533.36. This is considerably higher. The newstandard deviation is 850

√var[S5] = 65.486418 or £65,486.42 which is twice the previous amount. So there is

greater probability of covering the liability of £1,000,000, but there is also a much higher probability of a largeshortfall.

14. Let S2 denote the amount after 2 years. Then S2 = 10,000(1 + I1)(1 + I2) where

I1 =

0.03 with probability 1/3;0.04 with probability 1/3;0.06 with probability 1/3.

and I2 =

0.05 with probability 0.7;0.04 with probability 0.3.

Using the independence of I1 and I2 gives E[S2] = 10,000(1 + E[I1])(1 + E[I2]) = 10,000(1 + 0.13/3)(1.047) =10,923.7. (ii) E[S2

2 ] = 108E[(1 +I1)2]E[(1 +I2)2] = 108[1 + 2× 0.13/3 + 0.0061/3

][1 + 2× 0.047 + 0.00223] =

108 × 1.193465601. Hence var[S2] = 108(1.193465601− 1.092372) = 19,338.41

15. (i) S10 = 1,000Π10j=1(1 + Ij). Now E[1 + I1] = 1.06. Hence E[S10] = 1,000× 1.0610 = 1,790.85.

(ii) Now E[(1 + I1)2] = 1 + 0.12 + 0.00392 = 1.12392. Hence E[S210] = 106 × 1.1239210 and so var[S10] =

106 × [1.1239210 − 1.0620] = 9,145.60. Hence the standard deviation is 95.6326.(iii) (a) Clearly E[I1] is unchanged. Hence E[S10] is unchanged. The variance of I1 is smaller. Hence the standarddeviation of S10 will become smaller. (b) Clearly both E[S10] and the standard deviation of S10 will become larger.

16. The cash flow is as follows:Time 0 1 2 3 . . . 9 10Cash Flow (thousands) 150 20 20 20 . . . 20 20

HenceA(10) = 150(1+i1) · · · (1+i10)+20(1+i2) · · · (1+i10)+· · ·+20 where the 1+ij are i.i.d. lognormal(µ = 0.07, σ2 =0.006). Let α = eµ+ 1

2σ2

= e0.073. Then E[A(10)] = 150α10 +∑0j=9 20αj = 150α10 +20(α10−1)/(α−1) = 595.18471


giving the answer £595,184.71.(ii) Let x denote the required amount. We require P[x(1+i1) · · · (1+i10) ≥ 600] = 0.99. NowZ = (1+i1) · · · (1+i10) ∼lognormal(10µ, 10σ2) and hence lnZ ∼ N (10µ, 10σ2). So we require 0.99 = P[xZ ≥ 600] = P[lnZ ≥ ln 600 −lnx]. Hence 0.99 = P[(lnZ−10µ)/

√10σ2 ≥ (ln 600− lnx−10µ)/

√10σ2] where (lnZ−10µ)/

√10σ2 ∼ N (0, 1).

Hence (ln 600− lnx− 10µ)/√

10σ2 = −2.326 348 or lnx = ln 600− 10µ + 2.326 348√

10σ2 giving £526,771.15.

17. (i) Sn = (1 + i1) · · · (1 + in) where E[1 + it] = 1 + j, var[1 + it] = s2 and hence E[(1 + it)2] = s2 + (1 + j)2.Hence E[Sn] = (1 + j)n and E[S2

n] = [s2 + (1 + j)2]n. Hence var[Sn] = [s2 + (1 + j)2]n − (1 + j)2n.(ii) Now ln(1 + it) ∼ N (µ = 0.08, σ = 0.04). Hence lnS16 ∼ N (16µ, 16σ2).Hence Pr[1000S16 > 4250] = Pr[S16 > 4.25] = Pr[lnS16 > ln 4.25] = Pr[(lnS16 − 16µ)/(4σ) > (ln 4.25 −16µ)/(4σ)] = 1− Φ((ln 4.25− 16µ)/(4σ)) = 1− Φ((ln 4.25− 1.28)/0.16) = 1− Φ(1.0432436) = 0.1485 by usinginterpolation on the tables.

18. (i) Sn = (1 + i1) · · · (1 + in) where i1, . . . , in are i.i.d. with mean j and variance s2. HenceE[Sn] = (1 + j)n and E[S2

n] = E[(1 + i1)2]n = [var(1 + i1) + E(1 + i1)2]n = [s2 + (1 + j)2]n. Hence var(Sn) =[s2 + (1 + j)2]n − (1 + j)2n.(ii) Suppose j = 0.06 and s = 0.01. Hence s2 = 0.0001 and Z = ln(1 + i) ∼ N (µ, σ2).(a) E[1 + i] = eµ+ 1

2σ2

= 1.06 and s2 = 0.0001 = var[1 + i] = e2µ+σ2(eσ

2 − 1). Hence eσ2

= 0.0001/1.062 + 1 and soσ2 = 0.00008996 and eµ = 1.059928 and µ = 0.05822.(b) S12 = (1 + i1) · · · (1 + i12). Hence lnS12 ∼ N (12µ, 12σ2) = N (0.69869, 0.0010679).(c) Pr[S12 > 2] = Pr[lnS12 > ln 2] = Pr[(lnS12 − 0.69869)/σ > (ln 2− 0.69869)/0.032679] = 1− Φ(−0.1696) =Φ(0.1696) = 0.5636 + 0.96× 0.0039 = 0.567

19. Let W10 = (1 + i1) · · · (1 + i10) where i1, i2, . . . , i10 are i.i.d. with expectation 0.07 and standard deviation 0.09. NowE[(1 + i1)2] = 1 + 2× 0.07 + 0.092 + 0.072 = 1.153.(i) Hence we are interested in S10 = 1000W10. Now E[S10] = 1000E[W10] = 1000 × 1.0710 = 1967.15. UsingE[W 2

10] = 1.15310 gives the standard deviation of S10 to be 1000√

var[W10] = 1000√

1.15310 − 1.0720 = 531.66.(ii) We need the distribution of S10. Now ln(1 + i1) has a lognormal distribution. Define µ and σ by ln(1 + i1) ∼N (µ, σ2). Hence eµ+ 1

2σ2

= E[1 + i1] = 1.07 and e2µ+2σ2= E[(1 + i1)2] = 1.153. Hence eσ

2= 1.153/1.072 and so

σ2 = 0.00705. Hence eµ = 1.072/√

1.153 and so µ = 0.064134. It follows that ln(W10) ∼ N (10µ, 10σ2).We want P[S10 < 0.5E[S10]] = P[W10 < 0.5E[W10]] = P[ln(W10) < ln(0.5 × 1.0710)] = Φ[(ln(0.5 × 1.0710) −0.64134)/

√0.0705] = Φ(−2.4777974) = 0.00661.

(iii) P[1200W10 < 1400] = P[W10 < 7/6] = P[ln(W10) < ln(7/6)] = Φ[(ln(7/6) − 0.64134)/√

0.0705] =Φ[−1.835] = 0.0333.

20. Suppose invest £x at time 0. Then value at time 5 is A(5) = xS5 where S5 = R1R2R3R4R5 and R1, R2, . . . ,R5 arei.i.d. lognormal with expectation 1.04 and variance 0.02. Using the method of example 5.3b gives lnRk ∼ N (µ, σ2)where σ2 = ln(1 + 0.02/1.042) = 0.0183222 and µ = ln(1.042/

√0.02 + 1.042) = 0.0300596. Hence lnS5 ∼

N (5µ, 5σ2) or lnS5 ∼ N (0.150298, 0.091611).(i) We want x with 0.01 = P[xS5 < 5,000] = P[lnS5 < ln 5,000 − lnx] = P[(lnS5 − 0.150298)/

√0.091611 <

(ln 5,000 − lnx − 0.150298)/√

0.091611]. Hence (ln 5,000 − lnx − 5µ)/√

5σ2 = Φ−1(0.01) and hence x =exp

[ln 5,000− 5µ−

√5σ2Φ−1(0.1)

]= 8699.476176 or £8,699.48.

(ii) This amount is substantially more than £5,000, the size of the liability. The problem is that the returns haveexpectation 1.04, just larger than 1 but the variance is 0.02. Hence the variance is relatively large and there is asubstantial probability that the value of the return is less than 1.

21. (i) Now Sn = Πnk=1(1+ik) where i1, . . . , in are i.i.d. with mean j and standard deviation s. (a) Using independence

gives E[Sn] = Πnk=1(1 + E[ik]) = (1 + j)n. (b) Using independence again gives E[S2

n] = Πnk=1E[(1 + ik)2] =

Πnk=1(1 + 2j + E[i2k]) = (1 + 2j + s2 + j2)n. Hence var[Sn] = E[S2

n]− E[Sn]2 = (1 + 2j + s2 + j2)n − (1 + j)2n.(ii)(a) Now j = E[ik] = (i1 + i2)/2. Also E[i2k] = (i21 + i22)/2. Hence var[ik] = (i21 + i22)/2− (i1 + i2)2/4 = (i21 + i22 −2i1i2)/4 = (i1−i2)2/4. (b) Now (1+j)25 = 5.5; hence j = 0.07056862. Also (1+2j+s2 +j2)25− (1+j)50 = 0.52 =0.25. Hence (1+2j+s2 +j2)25 = 0.25+5.52 = 30.5. Hence s2 = 0.00037738. It follows that i1 +i2 = 2j = 0.14113724and i1 − i2 = 2s = 0.03885239. Hence i1 = 0.0899948 and i2 = 0.0511424 or 8.99948% and 5.11424%.

22. Suppose the effective interest rate is i per annum. Then the accumulated value at time 10 is

S(10) = 800[(1 + i) + (1 + i)2 + · · · + (1 + i)10] = 800(1 + i)

(1 + i)10 − 1i

We are given that

i =

0.02 with probability 0.250.04 with probability 0.550.07 with probability 0.2

and hence S(10) =

8934.97233579 with probability 0.259989.0811263 with probability 0.5511826.8794549 with probability 0.2

Hence E[S(10)] = 10093.1135944 or £10,093.11. (b) This is clearly 0.2.


23. (i) Let ik denote the effective rate of interest per annum in year k. Then S10 = (1+i1) · · · (1+i10). Using independencegives E[S10] = 1.0710 = 1.967151.(ii) Using independence again gives E[S2

10] = Π10k=1E[(1 + ik)2] = (1 + 2 × 0.07 + 0.016 + 0.072)10 and var[S10] =

E[S210]− E[S10]2 = (1 + 2× 0.07 + 0.016 + 0.072)10 − 1.0720 = 0.576097.

(iii) If 1,000 units are invested, then the expectation is 1,000E[S10] and the variance is 1,000,000var[S10].

24. Use units of £1,000.(i) Accumulated amount is A(3) = 80S3 where S3 = (1 + i1)(1 + i2)(1 + i3) and E[1 + i1] = 1.06, E[1 + i2] =34 × 1.07 + 1

4 × 1.05 = 1.065 and E[1 + i3] = 710 × 1.06 + 3

10 × 1.04 = 1.054. Using independence gives E[A(3)] =80× 1.06× 1.065× 1.054 = 95.18885 or £95,188.85.(ii) Now var[A(3)] = 6,400var[S3]. Using independence again gives E[S2

3 ] = E[(1+i1)2]E[(1+i2)2]E[(1+i3)2]. NowE[(1+i1)2] = 1

3 (1.042 +1.062 +1.082); E[(1+i2)2] = 34 1.072 + 1

4 1.052 = 1.1343, and E[(1+i3)2] = 710 1.062 + 3

10 1.042 =1.111. Hence E[S2

3 ] = 1.416304978. Hence var[A(3)] = 6,400(1.41630498−1.18986062) = 3.435073 or 3,435,073in units of £2.(iii) Now 80× 1.08× 1.07× 1.06 = 97.99488. For any other possibility we have A(3) < 97. Hence the probabilityis 1

3 ×34 ×

710 = 7

40 = 0.175.

25. Let Y = 1 + it. Then E[Y ] = 1.06 and var[Y ] = 0.082. Now ln(Y ) ∼ N (µ, σ2). Hence eµ+ 12σ

2= 1.06 and

e2µ+σ2(eσ

2 − 1) = 0.082. Hence eσ2

= 1 + 0.082/1.062 leading to σ2 = 0.00567982. Also eµ = 1.062/√

0.082 + 1.062

leading to µ = 0.0554290.

Let S10 denote the value at time 10 of £1 at time 0. Then S10 =∏10j=1(1 + ij). Hence ln(S10) ∼ N (10u, 10σ2). Hence

2E[S10] = 2 exp(10µ + 10σ2/2) = 2 exp(0.55429)× (1 + 0.082/1.062)5 = 3.5816954 or £3.581695m.(ii) We want P[2S10 < 0.8 × 3.5816954] = P[S10 < 1.432678] where ln(S10 ∼ N (0.554290, 0.0567982). HenceP[S10 < 1.432678] = Φ((ln(1.432678)− 0.554290)/

√0.0567982) = Φ(−0.817143) = 0.20692.

26. (i) Option A. Let A1 = 100(1 + i1)(1 + i2) · · · (1 + i10). Then E[A1] = 100 × 1.05510 = £170.81 by independence.Also E[A2

1] = 1002(1 + 2E[i1] + E[i21]

)10= 1002

(1.11 + 0.072 + 0.0552

)10.

Hence var[A1]/1002 = (1.11 + 0.072 + 0.0552)10 − 1.05520 giving the standard deviation of £36.20.Option B. For this case

A2 =

100× 1.065 × 1.015 with probability 0.2;100× 1.065 × 1.035 with probability 0.3;100× 1.065 × 1.065 with probability 0.2;100× 1.065 × 1.085 with probability 0.3.

Hence E[A2] = 100×1.065×[0.2×1.015 +0.3×1.035 +0.2×1.065 +0.2×1.085] = £169.48. Also var[A2] = 1002×1.0610×(0.2×1.0110 +0.3×1.0310 +0.2×1.0610 +0.3×1.0810−(0.2×1.015 +0.3×1.035 +0.2×1.065 +0.3×1.085)2

giving a standard deviation of £21.62.(ii) Option A. Now ln(1 + it) ∼ N (µ, σ2). Hence eµ+ 1

2σ2

= E[1 + it] = 1.055 and e2µ+2σ2= E[(1 + it)2] =

1.11 + 0.072 + 0.0552. Hence eσ2

= (1.11 + 0.072 + 0.0552)/1.0552 giving σ2 = 0.00439275.Also eµ = 1.0552/

√1.11 + 0.072 + 0.0552 giving µ = 0.0513444.

Now ln(A1/100) ∼ N (10µ, 10σ2) = N (0.513444, 0.0439275). We want

P[A1 < 115] = P[ln(A1/100) < ln 1.15] = Φ[

ln 1.15− 0.513444√0.0439275

]= 0.0372989 or 0.0373.

Option B. . The smallest possible amount is 100 × 1.065 × 1.015 = 140.648853 which is greater than 115. So theprobability is 0.(iii) Both options have similar expectations, but option A has a much higher variance and so is riskier. There is amuch higher probability of a low return with option A.

27. (i) LetA denote the accumulated value. ThenA = (1+i1)(1+i2) · · · (1+i10). By independence, E[A] = Π10j=1E[1+ij] =

E[1 + i1] (E[1 + i2])9 = 1.05 × 1.049 = 1.494477 or £1,494,477. (ii) Now E[(1 + i1)2] = 1 + 0.1 + 0.0029 = 1.1029and E[(1 + i2)2] = 1 + 0.08 + 0.00184 = 1.08184. Hence E[A2] = E[(1 + i1)2]

(E[(1 + i2)2]

)9= 1.1029× 1.081849.

Hence var[A] = E[A2]− (E[(A)])2 = 0.00527622 giving a standard deviation of 0.0726376 or £72,638.(iii) The value of E[i2] is unchanged. Hence the value of E[A] will be unchanged. The value of var[i2] is increased—hence the value of the standard deviation of A will be increased.

28. Let α = E(1 + i) = eµ+ 12σ

2. Let S20 denote the required answer. Then S20 = 5000(1 + i1) · · · (1 + i20) + 5000(1 +

i2) · · · (1 + i20) + · · · + 5000(1 + i20) and hence E(S20) = 5,000(α20 + α19 + · · · + α) = 5,000α(α20 − 1)/(α − 1) =5,000α(e1.04− 1)/(α− 1) = 180498.925165 or £180,498.93. (ii) For this part S20 = (1 + i1) · · · (1 + i20). HenceE(S20) = e20µ+10σ2

= e1.04. We want P(S20 > e1.04) = P(lnS20 > 1.04) where lnS20 ∼ N (20µ, 20σ2) = N (1, 0.08).Hence P(S20 > e1.04) = 1− Φ(0.04/

√0.08) = 0.44376854 or 0.44.


29. (i) The expected annual interest rate is 0.04 × 0.3 + 0.06 × 0.7 = 0.54. If x denotes the required amount, thenx × 1.05420 = 200,000 and hence we get x = 69858.262239 or £69,858.26. (ii) The value is x × 1.0420 withprobability 0.09, and x × 1.0620 with probability 0.49 and x × 1.0410 × 1.0610 with probability 0.49. This hasexpectation 201336.5443921 or £201,336.54 giving the answer £1,336.54.(iii) Smallest amount is x×1.0420 = 153068.050218 or £153,068.05; largest amount is x×1.0620 = 224044.903673or £224,044.90. Hence the range is £224,044.90− £153,068.05 = 70,976.85.

30. (i) Let A = 10,000(1 + i1)(1 + i2) · · · (1 + i15) where E[ij] = 0.07. Hence E[A] = 10,000 × 1.0715 = 27590.3154or £27,590.32. (ii) Now E[i2j] = 0.00506. Hence E[(1 + ij)2] = 1.14506. Hence E[A2] = 108 × 1.1450615 andhence var[A] = 108(1.1450615−1.0730), and so the standard deviation is 104

√1.1450615 − 1.0730 = 1263.83669 or

1,263.837. (iii) (a) The mean, E[ij], is clearly the same. Hence E[A] is unchanged. However, there is less variabilityand so the standard deviation of A will be less. (b) The mean will be less because the money is invested for a shorterperiod. Also the standard deviation will be less because the range of possibilities for A will be less.

31. (i) Let S5 = 7.85(1 + i1)(1 + i2)(1 + i3)(1 + i4)(1 + i5). Hence E[S5] = 7.85[E(1 + i1)]5 = 7.85 × 1.0555 =10.259636050 or £10,259,636. Now E[(1 + i1)2] = 1 + 2 × 0.055 + 0.042 + 0.0552 = 1.114625. Hence var[S5] =7.852

[1.1146255 − 1.05510

]and stdev[S5] = 7.85

√1.1146255 − 1.05510 = 0.871061457 or £871,061.

(ii) In this case S5 = 7.85 × 1.045 = 9.550725284 or £9,550,725. So there is a certain loss of £449,274 with thepolicy of part (ii). The policy of part (i) has an expected profit but there is the possibility of a very large loss. Usingthe approximation of 95% of a distribution within ±2 standard deviations, implies a chance of 1 in 40 of a loss ofover £1,400,000.

32. (i) Now Sn = (1 + i1) · · · (1 + in). Using independence gives E[Sn] =∏nt=1 E[1 + it] = (1 + j)n. Similarly E[S2

n] =[E(1 + it)2

]n=[1 + 2j + s2 + j2

]nbecause E[i2t] = var[it] + E[it]2. Hence var[Sn] = (1 + 2j + j2 + s2)n− (1 + j)2n.

(ii)(a) See notes: σ2 = ln(1 + 0.122/1.042) = 0.013226 and eµ = 1.042/√

0.122 + 1.042 leading to µ = 0.032608.(b) Now 1 + it ∼ lognormal(µ, σ2). Hence ln(1 + it) ∼ N (µ, σ2) and (ln(1 + it)− µ) /σ ∼ N (0, 1). HenceP[1.06 < 1 + it < 1.08] = Φ

(ln(1.08)− µ)/σ

)− Φ

((ln(1.06)− µ)/σ

)= 0.06184493 or 0.062 or 6.2%. (iii) The

expectation of the return is 4%. So the probability it lies between 6% and 8% will be small; The reason it is so smallis a consequence of the shape of the lognormal distribution. Try the following plot of the density: plot( seq(0,2,by=0.01), dlnorm( seq(0,2, by=0.01), m=mu,s=sqrt(sigma1)), type="l").


1. Now 1 + i = (1 + 0.05/4)4. Hence 80 = K/(1 + i)1/4 = K/1.0125 and so K = 80× 1.0125 = 81.

2. We have S0 = 12. Let i denote effective risk-free interest rate per annum and let K denote the forward price. ThenK

(1 + i)10/12 = 12− 1.5(1 + i)3/12 −

1.5(1 + i)6/12 −

1.5(1 + i)9/12

Now 1 + i = 1.032. Hence K = 12× 1.035/3 − 1.5× 1.037/6 − 1.5× 1.034/6 − 1.5× 1.031/6 = 8.01609 or £8.02.

3. (i) See page 131. (ii) Now Ke−0.07×7/12 = 60 − 2e−0.07×3/12 − 2e−0.07×6/12 which gives K = 60e0.49/12 −2e0.28/12 − 2e0.07/12 = 58.4418.

4. Now £1.20 invested for 91 days at 5% p.a. effective will accumulate to 1.20 × 1.0591/365 = 1.2147. No arbitrageimplies that the forward price to be paid in 91 days must be £1.2147.

5. (i) A forward contract is a legally binding agreement to buy or sell an agreed quantity of an asset at an agreed priceat an agreed time in the future. If the contract specifies that A will buy the asset from B at some time in the future,then A holds a long forward position and B holds a short forward position.(ii) Using time 0 values gives Ke−δT = S0e

−rT − ce−δt1 . Hence Ke−0.05/4 = 150e−0.03/4 − 30e−0.05/6 and henceK = 150e0.02/4 − 30e0.05/12 = 120.62661.

6. Let K denote the forward price. HenceK

1.045= 6− 0.3

1.041/2 −0.3

1.045This gives K = 6.27− 0.3× 1.045/1.041/2 − 0.3 = 5.662588.

7. (i) A forward contract is a legally binding contract to buy (or sell) an agreed quantity of an asset at an agreed priceat an agreed time in the future. The contract is usually tailor-made between two financial institutions or between afinancial institution and a client. Thus futures contracts are over-the-counter or OTC. Such contracts are not normallytraded on an exchange.Settlement of a forward contract occurs entirely at maturity and then the asset is normally delivered by the seller tothe buyer. Forward contracts are subject to credit risk—the risk of default.The party agreeing to sell the asset is said to hold a “short forward position” and the buyer is said to hold a “longforward position.”(ii) Original contract: Price S0 = 96. Hence K = S0 × 1.0410 − 7 × 1.04. Hence the agreed forward price wasK = 134.82.


Either: position now is that the price is S0 = 148. Hence K = S0 × 1.043 − 7× 1.04. Hence the new forward priceshould be K = 159.20. This gives a difference of 159.20− 134.82 = 24.38 or 24.28/1.043 = 21.67 in today’s terms.Or: the value of the contract to its owner who has the right to buy the asset in 3 years’ time for the price 134.82 is148− 7/1.042 − 134.82/1.043 = 21.67.

8. (i) See page 131. (ii) Let K denote the forward price, α = 1.01 and β = 1.015. The dividends are as follows:Time 0 0.25 0.5 0.75 1 1.25 1.5 1.75 2 2.25 2.5 2.75 3Dividend 0 0.2α 0.2α2 0.2α3 0.2α4 0.2α5 0.2α6 0.2α7 0.2α8 0.2α8β 0.2α8β2 0.2α8β3 0.2α8β4

Hence

Ke−0.15 = 4.5−8∑k=1

0.2αke−0.05k/4 −4∑k=1

0.2α8βke−0.1−0.05k/4

= 4.5− 0.28∑k=1

xk1 − 0.2α8e−0.14∑k=1

xk2 = 4.5− 0.2x11− x8

1

1− x1− 0.2α8e−0.1x2

1− x42

1− x2

where x1 = αe−0.05/4 = 0.99745358 and x2 = βe−0.05/4 = 1.00239147. Hence K = 2.47433598.

9. (i)(a) Accumulation is 100 exp(∫ 15

5 δ(t) dt)

= 100 exp(0.25) exp(

0.08t + 0.003t2

2

∣∣∣15

10

)= 100 exp(0.25) exp(0.4 +

0.1875) = 100 exp(0.25) exp(0.5875) = 231.058329.

(b) Accumulation is 100 exp(∫ 14

5 δ(t) dt)

== 100 exp(0.25) exp(

0.08t + 0.003t2

2

∣∣∣14

10

)= 100 exp(0.25) exp(0.32 +

0.144) = 100 exp(0.25) exp(0.464) = 204.214352.(c) Accumulation is 100 exp

(∫ 1514 δ(t) dt

)= 100 exp(0.5875− 0.464) = 113.145000.

(d) We want δ with 100e10δ = 231.058329 and hence δ = 0.083750.(ii) Now ν(t) = exp(−

∫ t0 δ(u) du). Hence for 0 ≤ t ≤ 5 we have ν(t) = exp(−0.05t). We need∫ 5

0ρ(t)ν(t) dt =

∫ 5

0100e0.01te−0.05t dt = 100

∫ 5

0e−0.04t dt = 100

e−0.04t

−0.04

∣∣∣∣50

= 2500(1− e−0.2) = 453.173117

(iii) Let K denote the forward price. Then Kν(1) = 300 − 7ν(0.5) where ν(t) is given above. Hence K =300 exp(0.05)− 7 exp(0.025) = 308.204123.

10. Use £100 nominal. Let K denote the forward price.Investor has a short position. This means he must supply the bond at time t = 1 for the price of £98. He receives acoupon of 2.5 at time t = 0.5 and a coupon of 2.5 at time t = 1. So the time t = 0 value of these three receipts is

98e−0.052 + 2.5e−0.023 + 2.5e−0.052 = 97.850707The current price is £95. Hence the value to the investor is £97.850707− £95 = £2.850707. This was for £100 nom-inal. For £1 million, the value is 10,000× £2.850707 = £28,507.07.

11. (i) This theory asserts that the relative attraction of long and short term bonds depends on the expectations of futuremovements in interest rates. If investors expect interest rates to fall, then long term bonds will become more attrac-tive; this will cause the yields on long term bonds to fall and lead to a decreasing yield curve. Similarly, if investorsexpect interest rates to rise, then long term bonds will become less attractive and so lead to an increasing yield curve.

(ii)(a) Clearly i1 = 0.08 or 8%. Now (1 + i2)2 = 1.08 × 1.07 and hence i2 = 0.07498837. Now (1 + i3)3 =1.08×1.07×1.06 and hence i3 = 0.06996885. Finally, (1+i4)4 = 1.08×1.07×1.06×1.05 and hence i4 = 0.06494131.(b) The price of the bond is

P = 5[

11 + i1

+1

(1 + i2)2 +1

(1 + i3)3 +1

(1 + i4)4

]+

100(1 + i4)4 = 94.67515038

So we need the gross redemption yield ofTime 0 1 2 3 4Cash flow −P 5 5 5 105

So we need to find i with P = 5a4 ,i + 100/(1 + i)4. To get a first approximation, use the usual method of spreadingthe capital gain over the coupons; this gives 5 + (100− P )/4 = 6.33 which suggests i lies between 6% and 7%.If i = 0.06, then 5a4 ,i + 100/(1 + i)4 − P = 1.8598.If i = 0.07, then 5a4 ,i + 100/(1 + i)4 − P = −1.44960.Using linear interpolation gives i = 0.06 + 0.01× 1.8598/(1.8598 + 1.4496) = 0.06562. or 6.56%.

(c) Let K denote the forward price. HenceK

(1 + i2)2 = 400− 41 + i1

and hence K = 1.08× 1.07× (400− 4/1.08) = 457.96


12. Suppose K denotes the non-arbitrage forward price. Then Ke−0.03/2 = 10− 0.5e−0.03/12. Hence K = 9.64484. Sothe forward price of £9.70 is too high. Hence a risk-free profit can be made as follows.

Time 0: borrow £10 and buy one share—the net cash flow is zero.Also enter into a forward contract to sell the share in 6 months’ time for £9.70.

Time 1 month: receive dividend of £0.50 and invest for 5 months.Time 6 months: dividend has grown to £0.5e0.03×5/12; also owe £10e0.03/2 to the bank; receive £9.70 for the share.So the return is £(9.70 + 0.5e0.03×5/12 − 10e0.03/2) = £(9.70−K) = £0.0551586.

13. (i) See page 131. (ii) The given information is summarised in the following table:t = 0 t = 1, market up t = 1, market down

cost of security A 20 25 15cost of security B 15 20 12

Now 25/20 = 15/12. Whatever the state at time t = 1, the prices of the shares are in the ratio 5 : 4. Hence for noarbitrage, the prices of the shares at time t = 0 must be in the ratio 5 : 4. Hence if the price of share A at time 0 is 20,then the price of share B must be 16.At 15, the price of B is too low: consider selling three A and buying four B at time 0. This leads to a profit at time 1whether the market rises or falls.

14. (i) Let K1 denote price of original forward contract of 12 years. ThenK1

1.0312 = 95− 51.037 −

61.039

Let K2 denote price of forward contract arranged now for 7 years. ThenK2

1.037 = 145− 51.032 −

61.034

Hence value of contract is K2 −K1 in 7 years’ time. In units of money now it isK2 −K1

1.037 = 145− 95× 1.035 = 34.86896

(ii) For this case:K1

1.0312 =95

1.0212

andK2

1.037 =145

1.027

Hence value of contract is K2 −K1 in 7 years’ time. In units of money now it isK2 −K1

1.037 =145

1.027 −95× 1.035

1.0212 =145× 1.025 − 95× 1.035

1.0212 = 39.39365

15. We have

900 =50

1.051/2 +50

1.06+

K

1.06This leads to K = 900× 1.06− 50× 1.06/1.051/2 − 50 = 852.2773 or 852p.

16. We have, in pence,

1000 =50

1.051/12 +50

1.057/12 +K

1.0511/12

Hence K = 1.054/12(1000× 1.057/12 − 50× 1.056/12 − 50) = 942.84489 or 9.43.

17. (i) Compare (A): buy asset at price B and (B): buy forward at price K and deposit Ke−δT in risk free asset. Attime T , both lead to same value—ownership of the asset. Hence, by no arbitrage, Ke−δT = B.(ii) Let K denote the price of the forward contract. Then

200− 101.02

=K

1.022

Hence K = 200× 1.022 − 10× 1.02 = 197.88 or £197.88.

18. Note: this is a poorly phrased question! (i) See page 131. (ii) Let t = 0 denote the time four years ago and letK0 denote the price of the initial forward contract at that time. The price of the security was £7.20 at t = 0 withdividends of £1.20 at t = 5, 6, 7, 8 and 9. So we are comparing the following 2 cash flows:

Time 0 1 2 3 4 5 6 7 8 9c1 −7.20 0 0 0 0 1.20 1.20 1.20 1.20 1.20c2 −K0

Hence K0 = 7.20− 1.2a5 ,0.025/1.0254. Let K1 denote the price of a forward contract at t = 4. Then K1 = 10.45−1.2a5 ,0.025. Hence the value of the original contract at t = 4 is K1 − 1.0254K0 = 10.45− 7.20× 1.0254 = 2.502547or £2.50.


(iii) Assume 1 unit of the security grows to 1.03t units of the security over the interval (0, t). Then the price ofthe original forward contract at t = 0 is 7.2/1.039 which has value 7.2 × 1.0254/1.039 at t = 4. The price ofa new forward contract at t = 4 is 10.45/1.035. Hence the value of the original forward contract at t = 4 is10.45/1.035 − 7.2× 1.0254/1.039 = 2.923201 or £2.92.

19. Let ν = 1/1.042.Date 1/9/12 1/12/12 1/3/13 1/6/13 1/7/13

Cash flow: −10 1 1 1 K

So 10 = ν1/4 + ν1/2 + ν3/4 +Kν10/12. Hence K = 7.595645867212 or £7.5956.(ii) The forward price is the price quoted today (1/9/12) for buying 1 share on 1/7/13. Whether or not the share priceincreases or decreases between 1/9/12 and 1/7/13 does not affect the forward price and indeed this information isunknown when the cost of the forward contract is determined. Owning a forward contract on 1/9/12 is like owningthe share on 1/9/12 except that no dividends before 1/7/13 are received and there is no need to pay the seller of theshare until 1/7/13.

20. (i)(a) A futures contract is a legally binding contract to buy or sell a specified quantity of an asset at a specified priceat a specified time in the future. An option gives the right but not the obligation to buy or sell a specified quantity ofan asset at a specified price at a specified time in the future.(b) A call option gives the owner the right but not the obligation to buy a specified asset for a specified price at somespecified time in the future. A put option gives the owner the right but not the obligation to sell a specified asset fora specified price at some specified time in the future.(ii) We are comparing the following 2 cash flows:

Time 1/4/2013 30/9/2013 31/3/2014c1 −10.50 1.10 1.10c2 −K

Assuming no arbitrage, we must have NPV(c1) = NPV(c2). Hence

−10.50 +1.1

1.0225+

1.11.0252 =

−K1.0252

Hence K = 8.801306 or £8.80.

21. Let K denote the forward price in pence. Then

180 =10

1.041/2 +K

1.043/4

Hence K = 1.041/4[180× 1.041/2 − 10

]= 175.274906066 or 175.275p.

22. (i) See page 131. (ii) See section 2.6 on page 136. Equating time 0 values givesKe−0.09×0.75 = 6e−0.035×0.75 and hence K = 6e(0.09−0.035)×0.75 = 6.2526756 or £6.25268.

(iii) Suppose that at time 0, the investor borrows the amount £6e−0.035×0.75 for 9 months and uses this to buy the stock;he reinvests the dividends in purchasing further units of the stock. Hence at time 0, the investor has e−0.035×0.75 unitsof the stock which grow to 1 unit after 9 months. Suppose further that at time 0, he enters into a forward contract tosell one unit of the stock at time 9 months for the amount £6.30.After 9 months, the investor will owe £6e−0.035×0.75 × e0.09×0.75 = £6e(0.09−0.035)×0.75 = £6.25268. Hence he makesa profit of £6.30− £6.25268.

23. Let δ = 0.05, the force of interest. The NPV at 1 October 2015 of the dividend at 1 November 2015 is 0.5e−δ/12

and the NPV of the dividend at 1 May 2016 is 0.5e−7δ/12. Finally, the NPV at 1 October 2015 of the amount K, theprice of the forward contract received at 1 July 2016 is Ke−9δ/12. Hence 10 = 0.5e−δ/12 + 0.5e−7δ/12 + Ke−9δ/12

giving K = 9.360988337958 or £9.36099. (ii) The investor has 2 choices: buy the share now, which costs £10 oragree now to a forward contract to buy the share in 9 months’ time. The price of this forward contract is fixed nowand so cannot depend on the price of the share in 9 months’ time which is unknown at this time.

Colophon. C denotes correction and A denotes addition.

2015/03 C: p 96. Question 30, exercises 5.5. C: pp 176–177. Answer to question 23, exercises 5.5.2016/01 A: index of notation, pdf links2016/02 A: September 2015 questions2016/03 A: section on Makeham’s formula

2007: April, September2008: April, September2009:2010:2011:2012: April, September2013: April, September2014:2015: April, September2016:

Page 210 Colophon Sep 27, 2016(9:51) ST334 Actuarial Methods c©R.J. Reed

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